how to find ph of nano3 given molarity

The mass (in grams) of water needed to be circulating each hour in order to keep the reactor at or below 95°C is 627390 g

The following data were obtained from the question:

The mass of water needed can be obtained as illustrated below:

Q = MCΔT

210000000 = M × 4.184 × 80

210000000 = M × 334.72

Divide both sides by 334.72

M = 210000000 / 334.72

M = 627390 g

Thus, we can conclude that the mass of the water needed is 627390 g

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We need 448 mL (or 0.448 L) of 0.130M HCl to react completely with 1.51g of Al(OH)3.

To solve this problem, we first need to write the balanced chemical equation for the reaction between hydrochloric acid (HCl) and aluminum hydroxide (Al(OH)3):

2Al(OH)3 + 6HCl → 2AlCl3 + 6H2O

From the equation, we can see that 2 moles of Al(OH)3 react with 6 moles of HCl. Therefore, we need to determine how many moles of Al(OH)3 we have in 1.51g:

1.51g Al(OH)3 × (1 mol Al(OH)3/78.0g Al(OH)3) = 0.0194 mol Al(OH)3

Since we need 6 moles of HCl to react with 2 moles of Al(OH)3, we need 3 moles of HCl to react with 1 mole of Al(OH)3. Thus, we need:

0.0194 mol Al(OH)3 × (3 mol HCl/1 mol Al(OH)3) = 0.0582 mol HCl

Now, we can use the concentration of the HCl solution (0.130M) to calculate the volume of solution we need:

0.0582 mol HCl ÷ 0.130 mol/L HCl = 0.448 L = 448 mL

Therefore, we need 448 mL (or 0.448 L) of 0.130M HCl to react completely with 1.51g of Al(OH)3.

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The pH of the buffer after 0.041 mol of Ba(OH)2 is added is 4.59. So, the pH after adding 0.041 mol of Ba(OH)2 to the solution is approximately 4.35.

To solve this problem, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA])
;where pH is the current pH of the buffer, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

First, we need to find the pKa of the acid. We can do this by using the concentration of the acid and its conjugate base:

pKa = pH + log([A-]/[HA])

pKa = 3.84 + log(0.33/0.25)

pKa = 3.84 + 0.16

pKa = 4.00

Now, we can use the Henderson-Hasselbalch equation to find the new pH after the Ba(OH)2 is added. When Ba(OH)2 is added, it will react with the conjugate base (y-) to form more of the acid (hy). This will shift the equilibrium and change the pH of the buffer.

We know that 0.041 mol of Ba(OH)2 is added to 0.67 L of the buffer. This means that the concentration of OH- ions in the buffer will increase by:

[OH-] = (0.041 mol / 0.67 L) / 2

[OH-] = 0.0307 M

Note that we divide by 2 because Ba(OH)2 produces two OH- ions for every mole.

Now, we can use the new concentration of the acid (hy) and the new concentration of the conjugate base (y-) to find the new pH:

pH = pKa + log([A-]/[HA])

[HA] = 0.25 mol/L - (0.041 mol / 0.67 L)

[HA] = 0.189 mol/L

[A-] = 0.33 mol/L + (0.041 mol / 0.67 L)

[A-] = 0.394 mol/L

pH = 4.00 + log(0.394/0.189)

pH = 4.00 + 0.59

pH = 4.59

Therefore, the pH of the buffer after 0.041 mol of Ba(OH)2 is added is 4.59.
To determine the new pH after adding 0.041 mol of Ba(OH)2 to the buffer solution, we first need to find the moles of H+ and Y- in the solution:

Moles of HY (acid) = 0.25 M * 0.67 L = 0.1675 mol
Moles of Y- (conjugate base) = 0.33 M * 0.67 L = 0.2211 mol

Ba(OH)2 is a strong base that dissociates completely, providing 2 moles of OH- ions per mole of Ba(OH)2. Thus:

Moles of OH- = 0.041 mol * 2 = 0.082 mol

Now, the OH- ions react with the acid (HY) to form water and the conjugate base (Y-). Since the amount of OH- ions is less than the amount of HY, the buffer capacity is not exceeded.

Moles of HY remaining = 0.1675 mol - 0.082 mol = 0.0855 mol
Moles of Y- after reaction = 0.2211 mol + 0.082 mol = 0.3031 mol

Now, we can use the Henderson-Hasselbalch equation to find the new pH:

pH = pKa + log([Y-] / [HY])

The initial pH is given as 3.84. We can use this to find the pKa:

pKa = pH - log([Y-] / [HY]) = 3.84 - log(0.33 / 0.25) ≈ 3.6

Now, we can find the new pH:

New pH = pKa + log([new Y-] / [new HY]) = 3.6 + log(0.3031 / 0.0855) ≈ 4.35

So, the pH after adding 0.041 mol of Ba(OH)2 to the solution is approximately 4.35.

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The main difference between molten (liquid) steel and solid steel is the amount of energy present in the form of heat. Molten steel has more heat energy, causing it to be in a liquid state, while solid steel has less heat energy, making it solid.

A refrigerator cools food and drinks by removing heat energy from the items inside and releasing it into the surrounding environment. This is accomplished through a process that involves a refrigeration cycle with a coolant or refrigerant. Here's a step-by-step explanation:

1. The refrigerator contains a refrigerant that begins as a cool, low-pressure gas. This refrigerant flows through the evaporator coils located inside the refrigerator.

2. As the refrigerant flows through the evaporator coils, it absorbs heat from the items inside the refrigerator, causing the refrigerant to heat up and evaporate into a gas.

3. This heated gas then moves to the compressor, which compresses the gas, increasing its temperature and pressure.

4. The hot, high-pressure refrigerant gas travels through the condenser coils located outside the refrigerator, where it releases the absorbed heat into the surrounding environment. This causes the gas to condense back into a liquid.

5. The now-cooled liquid refrigerant moves through an expansion valve, reducing its pressure and temperature. This prepares the refrigerant to re-enter the evaporator coils and repeat the cycle.

This continuous cycle of absorbing heat from the inside of the refrigerator and releasing it into the room allows the refrigerator to effectively cool food and drinks.

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The concentration of K+ ions in the solution is 0.15 M.

The dissolution reaction for[tex]KMnO_4[/tex] is[tex]KMnO_4[/tex](s) --> [tex]K^+[/tex](aq) +[tex]MnO_4^-[/tex](aq). This means that for every one mole of KMnO4 that dissolves, one mole of K+ ions and one mole of MnO4- ions are produced.

If a solution is made that is 0.15 M in [tex]KMnO_4[/tex], the concentration of K+ ions would also be 0.15 M, because the molar ratio of [tex]K^+[/tex] ions to [tex]KMnO_4[/tex] is 1:1.

Therefore, the concentration of [tex]K^+[/tex] ions in the solution is 0.15 M.

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So the pH at the equivalence point of this titration is 3.18.

The pH at the equivalence point of a titration between a weak acid and a strong base (such as KOH) can be calculated using the formula:

pH = pKa + log([base]/[acid])

In this case, the weak acid has a Ka value of 6.6 x 10^-4, which means its pKa is 3.18 (calculated as -log(6.6 x 10^-4)). At the equivalence point, the moles of KOH added will be equal to the moles of the weak acid initially present.

To find the moles of weak acid initially present, we can use the formula:

moles = concentration x volume

So:

moles of weak acid = 0.10 M x 0.0250 L = 0.0025 moles

At the equivalence point, this will react completely with the KOH, which has a concentration of 0.10 M:

0.0025 moles of weak acid = 0.10 M x volume of KOH added

Solving for volume of KOH added:

volume of KOH added = 0.0025 moles / 0.10 M = 0.0250 L

So the total volume of the solution at the equivalence point is 0.0500 L (25.0 mL of weak acid + 25.0 mL of KOH).

Now we can use the equation above to find the pH:

pH = 3.18 + log([0.10 M]/[0.10 M])

Simplifying:

pH = 3.18 + log(1)

pH = 3.18

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Answer:

One mole of hydrogen gas in 0.5 moles of oxygen gas is produced from one mole of liquid water collected in a separate 10 L container at 1 atm

Explanation:

Correct answer: The final chloride ion concentration is 1.14 M.

To find the final chloride ion concentration, first, we need to calculate the moles of ZnCl2 and then determine the concentration of the solution.
1. Calculate moles of ZnCl2:
moles = mass / molar mass
moles = 65 g / 136.29 g/mol
moles = 0.4765 mol
2. Convert volume from mL to L:
volume = 837 mL / 1000
volume = 0.837 L
3. Calculate concentration of ZnCl2 solution:
concentration = moles / volume
concentration = 0.4765 mol / 0.837 L
concentration = 0.57 M
Since there are 2 chloride ions (Cl-) for each ZnCl2 molecule, the final chloride ion concentration is double the concentration of ZnCl2:
Cl- ion concentration = 2 * 0.57 M
Cl- ion concentration = 1.14 M

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H₂CO₃ is an acid that gives a proton to H₂O, which accepts the proton as a base. H₂O's conjugate acid is H₃O⁺, and H₂CO₃'s conjugate base is HCO₃⁻. H₂O/H₃O⁺ and H₂CO₃/HCO₃⁻are the conjugate acid-base pair in this process.

Two chemical species are referred to as an acid-base pair when a proton (H+) from one species is transferred to the other. The species that accepts a proton is known as the base, while the species that provides a proton is known as the acid.

b. H₂O functions as an acid by giving a proton to a different H₂O molecule, which accepts the proton and functions as a base. H₃O⁺ is the conjugate acid of H₂O, whereas OH⁻ is the conjugate base of H₂O. H₂O/H₃O⁺ and H₂O/OH⁻ are the conjugate acid-base pair in this process.

c. H₂S functions as an acid by giving a proton to NH₃, which accepts the proton and functions as a base. NH₄⁺ is the conjugate base of H₂S, whereas NH₃'s conjugate acid is NH₄⁺. NH₃/NH₄⁺ and H₂S/HS⁻ are the conjugate acid-base pair in this reaction, respectively.

d. H₂PO₄⁻ functions as an acid by giving a proton to H₂O, while H₂O functions as a base by taking the proton. H₂O's conjugate acid is H₃O⁺, and H₂PO₄⁻'s conjugate base is HPO₄²⁻. H₂O/H₃O⁺ and H₂PO₄⁻/HPO₄²⁻ are thus the conjugate acid-base pair in this process.

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A. To find the moles of oxygen gas generated, we need to first calculate the moles of hydrogen peroxide in the 30.0 mL solution.

Mass of H2O2 = 4.7% of mass of solution
Mass of solution = 30.0 mL x 1.00 g/mL = 30.0 g
Mass of H2O2 = 0.047 x 30.0 g = 1.41 g

Now we can use the molar mass of hydrogen peroxide to calculate the moles:

Molar mass of H2O2 = 2(1.01) + 2(16.00) = 34.02 g/mol
Moles of H2O2 = 1.41 g / 34.02 g/mol = 0.0415 mol

According to the balanced equation, 2 moles of hydrogen peroxide will generate 1 mole of oxygen gas. Therefore:

Moles of O2 = 0.5 x 0.0415 mol = 0.0208 mol

So, if all the peroxide decomposed, 0.0208 moles of oxygen gas could be generated.

B. If the initial pressure was 0.985 atm and the moles of oxygen gas generated were 0.0208, we can use the ideal gas law to find the final pressure:

PV = nRT

We know the initial pressure (P), the number of moles of oxygen gas (n), and the temperature (T) remains constant. We can assume a value for volume (V) as the pressure change is what we are interested in.

Let's assume a volume of 100 mL:

V = 100 mL = 0.1 L

The gas constant (R) is 0.08206 L atm/mol K.

Plugging in these values:

0.985 atm x V = 0.0208 mol x 0.08206 L atm/mol K x T

Solving for T:

T = (0.985 atm x 0.1 L) / (0.0208 mol x 0.08206 L atm/mol K) = 59.4 K

Now we can use the ideal gas law again to find the final pressure:

PV = nRT

P = (nRT) / V = (0.0208 mol x 0.08206 L atm/mol K x 59.4 K) / 0.1 L = 0.101 atm

So, if the moles of oxygen gas generated were 0.0208 and the initial pressure was 0.985 atm, the final pressure would be 0.101 atm.

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After completing the titration, we can use the volumes of the acid and base solutions, as well as the concentration of the sodium hydroxide, to calculate the concentration of the unknown acid solution.

To determine the concentration of an unknown acid solution using acid-base titration, you can follow the steps provided in the procedure. Here's a step-by-step explanation:

1. Obtain approximately 80 mL of 0.1M sodium hydroxide (NaOH) solution in a labeled beaker. Record the actual concentration from the bottle.
2. Obtain around 25 mL of unknown acid in a second labeled beaker. Record whether the acid is monoprotic, diprotic, or tricrotic.
3. Fill the burette with sodium hydroxide solution, making sure to rinse it first with around 5-10 mL of NaOH to ensure accurate measurements.
4. Record the initial volume of NaOH in the burette to the correct number of significant figures.
5. Transfer 10.00 mL of the unknown acid into an Erlenmeyer flask using a volumetric pipette.
6. Add 3 drops of pH indicator (e.g., phenolphthalein) and a magnetic stirrer to the Erlenmeyer flask.
7. Place the Erlenmeyer flask under the burette on a hot plate (with the heat off) and turn on the stir knob to mix the solution.
8. Titrate the unknown acid with sodium hydroxide solution until the solution turns pink and remains pink, indicating the equivalence point has been reached.
9. Record the final volume of NaOH in the burette to the correct number of significant figures.
10. Repeat steps 4-7 if needed for additional trials.

After completing the titration, we can use the volumes of the acid and base solutions, as well as the concentration of the sodium hydroxide, to calculate the concentration of the unknown acid solution.

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The correct unit for heat capacity is: D) J/degree C (joules per degree Celsius)

Heat capacity is a thermodynamic property that measures the amount of heat energy required to change the temperature of a substance by one degree Celsius (or one Kelvin, which is equivalent in magnitude). Heat capacity is typically denoted by the symbol C and has units of J/°C (joules per degree Celsius) or J/K (joules per Kelvin).

The heat capacity of a substance is determined by the amount of energy it takes to increase the kinetic energy of its constituent particles, such as molecules or atoms, and thus increase its temperature. The higher the heat capacity of a substance, the more energy is required to achieve a given change in temperature. Conversely, substances with lower heat capacities require less energy to change temperature.

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standard enthalpies of formation: δh0f (co2 (g)) = -393.5 kj/mol δh0f (h2o (l)) = -285.8. KJ/mol,  the standard enthalpy of formation for methanol (CH3OH) is 965.1 kJ/mol.

To calculate the standard enthalpy of formation for methanol (CH3OH), we will use the following reaction:

CO (g) + 2 H2 (g) → CH3OH (l)

Now, we'll apply Hess's Law, which states that the enthalpy change of a reaction is the same, whether it occurs in one step or several steps. First, we need to find the enthalpy change for the reverse of the desired reaction:

CH3OH (l) → CO (g) + 2 H2 (g)

Next, let's find the enthalpy changes for the formation of CO2 and H2O:

C (s) + O2 (g) → CO2 (g); ΔH°f = -393.5 kJ/mol

2 H2 (g) + O2 (g) → 2 H2O (l); ΔH°f = -2 x (-285.8 kJ/mol) = -571.6 kJ/mol

Now, we need to find the enthalpy change for the reaction of CO2 and H2O to form CO and 2 H2:

CO2 (g) + 2 H2O (l) → CO (g) + 2 H2 (g) + O2 (g)

For this reaction:

ΔH° = ΔH°f (CO) + 2ΔH°f (H2) - ΔH°f (CO2) - 2ΔH°f (H2O)

Substitute the given values:

ΔH° = ΔH°f (CH3OH) - (-393.5 kJ/mol) - (-571.6 kJ/mol)

Now, we need to find ΔH°f (CH3OH):

ΔH°f (CH3OH) = ΔH° + 393.5 kJ/mol + 571.6 kJ/mol

ΔH°f (CH3OH) = 393.5 kJ/mol + 571.6 kJ/mol = 965.1 kJ/mol

Therefore, the standard enthalpy of formation for methanol (CH3OH) is 965.1 kJ/mol.

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The ranking of the following in order of increasing solubility in water, all containing only palmitic acid, is as follows: 1. Triacylglycerol, 2. Diacylglycerol and 3. Monoacylglycerol.

To rank the following in order of increasing solubility in water, we need to consider the polar and non-polar characteristics of each molecule.

1. Triacylglycerol: This molecule has three palmitic acid chains and is the most nonpolar of the three. Nonpolar molecules are generally less soluble in water, so triacylglycerol has the lowest solubility.

2. Diacylglycerol: This molecule has two palmitic acid chains, making it more polar than triacylglycerol but less polar than monoacylglycerol. Its solubility in water is between that of triacylglycerol and monoacylglycerol.

3. Monoacylglycerol: This molecule has one palmitic acid chain, making it the most polar of the three. Polar molecules tend to be more soluble in water, so monoacylglycerol has the highest solubility.

In conclusion, the order of increasing solubility in water is: triacylglycerol < diacylglycerol < monoacylglycerol.

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The solubility of iodine in water increases with increasing concentration of potassium iodide due to the shift in equilibrium towards the formation of more soluble triiodide ions.

Le Chatelier's principle states that when a system at equilibrium is subjected to a stress, the system will respond in a way that tends to counteract the stress.

In this case, the stress is the addition of potassium iodide, which provides more iodide ions, shifting the equilibrium towards the formation of more triiodide ions, which are more soluble in water than iodine. As a result, more iodine dissolves in water, increasing its solubility.

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The change in entropy when 86 g of diethyl ether melts at its melting point is approximately 11.81 J/(mol*K).

To calculate the change in entropy (ΔS) when 86 g of diethyl ether melts, we need to use the equation:
                                      ΔS = AfHf / T
Where AfHf is the heat of fusion (1854 kJ/mol) and T is the melting point temperature.

The melting point of diethyl ether is -116.3°C or 156.85 K (since we need to use Kelvin for temperature in this equation).
First, we need to calculate the number of moles of diethyl ether in 86 g:
86 g / 74.12 g/mol = 1.16 mol

Now, we can calculate the change in entropy:
ΔS = (1854 kJ/mol) / (156.85 K)
ΔS = 11.81 J/(mol*K)
Therefore, the change in entropy when 86 g of diethyl ether melts is 11.81 J/(mol*K).

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Answer:Most amphibians start their life cycle as eggs that are laid in the water. The eggs hatch into larvae, which are commonly referred to as tadpoles. Tadpoles spend their early life in the water, feeding on algae and other plant material. As they grow and develop, they undergo metamorphosis, which is a process where they transform into terrestrial adults. During metamorphosis, the tadpoles develop legs and lungs, and their tails are reabsorbed. Once the metamorphosis is complete, the young amphibians are able to live on land and breathe air. This life cycle is common to most amphibians, including frogs, toads, and salamanders.

Explanation:

Answer:

Amphibians typically lay eggs in water which hatch into larvae (aquatic juveniles). These larvae undergo metamorphosis and eventually emerge from the water as land-dwelling adults. Some species may also reproduce on land through a process called oviparity where they lay eggs directly onto land.

May I please have brainliest?

(a) The mass of the isotope is decreasing at a rate of 0.776 g/yr at that time.

(b) The half life of the isotope is approximately 4.39 years.

To answer this question, we need to use the radioactive decay formula:

N(t) = N₀ e^(-λt)

where N(t) is the amount of radioactive material at time t, N₀ is the initial amount of radioactive material, λ is the decay constant, and e is the mathematical constant approximately equal to 2.718.

(a) To find how quickly the mass of the isotope is decreasing at the given time, we need to find the rate of change of N(t) with respect to time:

dN/dt = -λN₀ e^(-λt)

We know that N₀ = 5.00 g and N(t) = 4.27 g after one year, so we can plug in these values and solve for λ:

4.27 = 5.00 e^(-λ*1)
e^(-λ) = 4.27/5.00
e^(-λ) = 0.854
-λ = ln(0.854)   (taking the natural logarithm of both sides)
λ ≈ 0.158 yr⁻¹

Now we can plug in λ and N₀ to find the rate of change of the mass of the isotope:

dN/dt = -λN₀ e^(-λt)
dN/dt = -(0.158 yr⁻¹)(5.00 g) e^(-0.155t)

At t = 1 year (since one year has passed since the initial measurement), we get:

dN/dt = -(0.158 yr⁻¹)(5.00 g) e^(-0.155)

dN/dt ≈ -0.676 g/yr

Therefore, the mass of the isotope is decreasing at a rate of approximately 0.776 g/yr at that time.

(b) To find the half-life of the isotope, we can use the formula:

t½ = ln(2)/λ

We already know the value of λ from part (a), so we can plug it in:

t½ = ln(2)/0.158 yr⁻¹

t½ ≈ 4.39 years

Therefore, the half-life of the isotope is approximately 4.39 years.

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it is acceptable to measure the Kl and HCI solutions used in titration with graduated cylinders because the concentration of these solutions is usually not critical to the experiment and does not require the precision of a buret or volumetric pipet.

Graduated cylinders are not as precise as burets or volumetric pipets because they typically have larger graduations and are not designed for precise measurements.
Regarding the procedural errors, failing to allow the volumetric pipet to drain completely when transferring the diluted bleaching solution to the Erlenmeyer flask would result in an incorrectly high calculated percent NaOCI in the commercial bleaching solution. This is because the solution would be more concentrated than intended.
Blowing the last drops of solution from the pipet into the volumetric flask when transferring commercial bleaching solution to the flask would also result in an incorrectly high calculated percent NaOCI. This is because the volume of solution in the flask would be greater than intended. it is acceptable to measure the Kl and HCI solutions used in titration Beginning a titration with an air bubble in the buret tip would result in an incorrectly low calculated percent NaOCI in the commercial bleaching solution. This is because the volume of Na_2S_2O_3 solution added would be less than intended due to the presence of the air bubble.

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1. Carefully measure out 5.47 mL of the 3.20 M [tex]MgCl_2[/tex] solution using a 100 mL graduated cylinder.

2. Transfer the 5.47 mL of [tex]MgCl_2[/tex] solution that was measured into a 250 mL volumetric flask.

3. To suitably dilute the [tex]MgCl_2[/tex] solution, add water to the volumetric flask up to the 250 mL mark.

4. Combine the flask's contents to get a consistent solution.

To prepare the desired solution, follow these steps:

1. Using a 100 mL graduated cylinder, carefully measure out 5.47 mL of the 3.20 M [tex]MgCl_2[/tex]  solution.
2. Pour the measured 5.47 mL of the [tex]MgCl_2[/tex]  solution into a 250 mL volumetric flask.
3. Fill the volumetric flask up to the 250 mL mark with water, ensuring the [tex]MgCl_2[/tex] solution is properly diluted.
4. Mix the contents of the flask to create a uniform solution.

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The statement is true. A molecule with polar bonds can either be polar or nonpolar depending on the arrangement of the polar bonds. Polar bonds are formed when two atoms with different electronegativities bond, resulting in an unequal sharing of electrons.

The more electronegative atom attracts the electrons towards itself, creating a partial negative charge while the other atom has a partial positive charge.

If the polar bonds in a molecule are arranged symmetrically, such that the partial charges on each atom cancel out, then the molecule is nonpolar.

However, if the polar bonds are arranged asymmetrically, such that the partial charges do not cancel out, then the molecule is polar.

For example, carbon dioxide (CO2) has two polar bonds between the carbon and oxygen atoms, but the molecule itself is nonpolar because the bonds are arranged symmetrically around the carbon atom, canceling out the partial charges.

On the other hand, water (H2O) also has polar bonds between the hydrogen and oxygen atoms, but the molecule is polar because the bonds are arranged asymmetrically, resulting in a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms.

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Part A: [tex]Na^2Se[/tex]
Part B: SrSe
Part C: [tex]BaCl^2[/tex]
Part D: [tex]Al^2O^3[/tex]

The formulas for these compounds using Lewis theory.

Part A: Na and Se
Lewis theory suggests that elements will combine to achieve a full valence electron shell. Sodium (Na) has one valence electron and Selenium (Se) has six valence electrons. Sodium will lose its one valence electron, and Selenium will gain two electrons to achieve a full valence shell.
Therefore, the formula for the compound formed between Na and Se is [tex]Na^2Se[/tex].

Part B: Sr and Se
Strontium (Sr) has two valence electrons, and Selenium (Se) has six valence electrons. Strontium will lose its two valence electrons, and Selenium will gain two electrons to achieve a full valence shell.
Therefore, the formula for the compound formed between Sr and Se is SrSe.

Part C: Ba and Cl
Barium (Ba) has two valence electrons, and Chlorine (Cl) has seven valence electrons. Barium will lose its two valence electrons, and each Chlorine atom will gain one electron to achieve a full valence shell.
Therefore, the formula for the compound formed between Ba and Cl is [tex]BaCl^2[/tex].

Part D: Al and O
Aluminum (Al) has three valence electrons, and Oxygen (O) has six valence electrons. Aluminum will lose its three valence electrons, and each Oxygen atom will gain two electrons to achieve a full valence shell.
Therefore, the formula for the compound formed between Al and O is [tex]Al^2O^3[/tex].

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The pH of the solution after adding 11.17 g of C2H5NH3Cl is 2.64.

To solve this problem, we need to use the concept of the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa or pKb of the acid or base present in the solution, respectively. In this case, we have a base (C2H5NH2) with a known pKb value of 3.19.

First, we need to find the concentration of C2H5NH2 in the initial solution. We can do this using the formula:

molarity = moles of solute / volume of solution (in liters)

Since we have 1000 ml (or 1 liter) of a 0.435 m solution of C2H5NH2, we can calculate the number of moles of C2H5NH2 as:

moles of C2H5NH2 = molarity x volume = 0.435 x 1 = 0.435 moles

Next, we need to find the concentration of C2H5NH3+ (the conjugate acid of C2H5NH2) after adding 11.17 g of C2H5NH3Cl. We can do this by first calculating the number of moles of C2H5NH3Cl added:

moles of C2H5NH3Cl = mass / molar mass = 11.17 g / 81.5446 g/mol = 0.1368 moles

Since C2H5NH3Cl is a salt that dissociates completely in water, it will produce an equal number of moles of C2H5NH3+ and Cl- ions. Therefore, the concentration of C2H5NH3+ after adding the salt is:

concentration of C2H5NH3+ = moles of C2H5NH3+ / volume of solution (in liters)

We need to find the volume of the final solution, which is the sum of the initial volume (1 liter) and the volume of the added salt. To find the volume of the added salt, we can use its density, which is 1.014 g/ml:

volume of added salt = mass of added salt / density = 11.17 g / 1.014 g/ml = 11 ml

Therefore, the total volume of the final solution is:

total volume = 1000 ml + 11 ml = 1011 ml = 1.011 liters

Now we can calculate the concentration of C2H5NH3+ as:

concentration of C2H5NH3+ = moles of C2H5NH3+ / volume of solution = 0.1368 moles / 1.011 liters = 0.135 mol/l

Finally, we can use the Henderson-Hasselbalch equation to find the pH of the solution:

pH = pKb + log([C2H5NH3+]/[C2H5NH2])

Substituting the values we have calculated, we get:

pH = 3.19 + log(0.135/0.435) = 3.19 - 0.55 = 2.64

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An enzyme called glycogenin has a number of roles in the manufacture and extension of glycogen, a glucose polymer that is used as a type of energy storage.

By activating the self-glycosylation of its own tyrosine residue, it first serves as a primer for glycogen extension and produces a short glucose polymer that can be extended by glycogen synthase. Second, glycogenin creates a new, extendable reducing end by cleaving alpha-1-4 glycosidic bonds in a developing glycogen chain.

Thirdly, it transfers a glucose residue from its own polymer to activate glucose-1-phosphate, transforming it to glucose-6-phosphate. Finally, glycogenin creates new branch points in the glycogen molecule, allowing for further glycogen synthesis, by cleaving alpha-1-6 glycosidic linkages.

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The best choice to produce a buffer with a pH greater than 7 is the mixture of 1.0 M NH3 and 1.0 M NH4+.

Here's a step-by-step explanation:

1. A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. It resists changes in pH when small amounts of acid or base are added.

2. For a pH greater than 7, we need a weak base and its conjugate acid. In the given options, NH3 (ammonia) is a weak base and NH4+ (ammonium ion) is its conjugate acid.

3. The mixture of 1.0 M NH3 and 0.1 M NH4+ would also produce a buffer, but it would be less effective because the concentrations of the weak base and its conjugate acid are not equal.

4. The mixture of 1.0 M CH3CO2H (acetic acid) and 1.0 M NaCH3CO2 (sodium acetate) forms a buffer, but with a pH less than 7, since CH3CO2H is a weak acid and CH3CO2- (acetate ion) is its conjugate base.

5. The mixture of 1.0 M CH3CO2H and 0.1 M NaCH3CO2 would also produce a buffer with a pH less than 7, for the same reasons mentioned in step 4.

In conclusion, the best choice to produce a buffer with a pH greater than 7 is the mixture of 1.0 M NH3 and 1.0 M NH4+, as it contains equal concentrations of a weak base and its conjugate acid.

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To calculate the molar heat of solution of KCN, we need to use the formula:

q = n∆H

where q is the heat absorbed or released during the process, n is the number of moles of KCN dissolved, and ∆H is the molar heat of solution.

We know that the temperature of the calorimeter decreases by 0.400 K when a 2.00 g sample of KCN is dissolved in water.

The total heat capacity of the calorimeter is given as 0.898 kJ⋅K−1. This means that the heat absorbed by the calorimeter can be calculated as:

q = C∆T

q = (0.898 kJ⋅K−1) × (0.400 K)

q = 0.3592 kJ

To calculate the number of moles of KCN dissolved, we need to use its molar mass which is 65.12 g⋅mol−1.

n = m/M

n = 2.00 g / 65.12 g⋅mol−1

n = 0.0307 mol

Substituting the values for q and n in the formula, we get:

0.3592 kJ = (0.0307 mol) × ∆H

Solving for ∆H, we get:

∆H = 11.70 kJ⋅mol−1

Therefore, the molar heat of solution of KCN is 11.70 kJ⋅mol−1.

This value indicates that KCN dissolves in water with the release of heat, which is an exothermic process.

This information can be useful in various chemical processes where the heat of solution plays an important role.

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The equilibrium constant for the Haber process at room temperature is 5.46 x 10^-4, and the standard emf of the process at room temperature is -0.477 V.

Involving the thermodynamic information in Supplement C, we can work out the standard Gibbs free energy change for the response and afterward utilize that worth to ascertain the balance steady and standard emf.

The standard Gibbs free energy change for the Haber cycle can be determined utilizing the condition:

ΔG° = ΣnΔG°f(products) - ΣmΔG°f(reactants),

where n and m are the stoichiometric coefficients of the items and reactants, and ΔG°f is the standard Gibbs free energy of development for each compound.

For the Haber cycle, we have:

ΔG° = 2ΔG°f(NH3) - ΔG°f(N2) - 3ΔG°f(H2)

Subbing the qualities from Index C, we get:

ΔG° = (2*(- 16.45 kJ/mol)) - (- 16.59 kJ/mol) - 3*(- 0.0 kJ/mol) = - 92.28 kJ/mol.

To ascertain the harmony steady at room temperature, we can utilize the relationship:

ΔG° = - RTln(K)

where R is the gas steady and T is the temperature in Kelvin.

Subbing the qualities, we get:

-92.28 kJ/mol = - (8.314 J/Kmol) * 298 K * ln(K)

Settling for K, we get:

K = [tex]e^(- 92.28 kJ/mol/(8.314 J/Kmol * 298 K))[/tex] = [tex]5.46 x 10^-4[/tex]

To ascertain the standard emf of the Haber interaction, we can utilize the relationship:

ΔG° = - nFE°

where n is the quantity of electrons moved in the response and F is Faraday's steady (96,485 C/mol).

For the Haber interaction, two electrons are moved, so n = 2. Subbing the qualities, we get:

-92.28 kJ/mol = - 2 * (96,485 C/mol) * E°

Settling for E°, we get:

E° = - 92.28 kJ/mol/(2 * 96,485 C/mol) = - 0.477 V

Hence, the balance consistent for the Haber cycle at room temperature is [tex]5.46 x 10^-4[/tex], and the standard emf of the interaction at room temperature is - 0.477 V. This implies that the response isn't unconstrained under standard circumstances, and a voltage of no less than 0.477 V should be applied to drive the response in the forward heading.

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Depressing the plunger all the way to the base when filling a Variable volume pipet can result in an unintentional extra volume, affecting the change in time and the resulting calculated rate.

understand that you want to know how the unintentional extra volume affects the change in time and calculated rate when using a variable volume pipet. Let's analyze the situation step-by-step:

1. A student accidentally depresses the plunger all the way to the base while filling the pipet, unintentionally adding an extra 2 to 3 mL of volume.

2. Due to this extra volume, the amount of liquid being transferred is higher than intended, which may lead to inaccurate results in experiments.

3. Since more liquid is being transferred, the time required to complete the transfer will also increase, leading to a change in time.

4. The calculated rate is usually determined by dividing the volume of liquid transferred by the time taken. In this case, both the volume and time have increased, which might lead to an inaccurate calculation of the rate.

depressing the plunger all the way to the base when filling a variable volume pipet can result in an unintentional extra volume, affecting the change in time and the resulting calculated rate.

This can lead to inaccurate results in experiments, so it's essential to properly use pipets to ensure precise measurements.

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Answer:

I don't know the answer I will answer you after 20 years

A chemical that is extremely volatile, flammable, and capable of forming explosive peroxides upon long-term contact with atmospheric oxygen is an ether, specifically diethyl ether.

Diethyl ether is highly volatile due to its low boiling point of 34.6°C (94.3°F), which means it can evaporate quickly at room temperature. This volatility allows it to form a flammable vapor-air mixture, making it a fire and explosion hazard. It can easily ignite in the presence of a spark, heat, or flame.

Moreover, diethyl ether can form explosive peroxides over time when exposed to atmospheric oxygen. Peroxides are chemical compounds containing a peroxide group (O-O) that are highly reactive and unstable. These peroxides can accumulate in storage containers or on the surface of the diethyl ether, making them even more dangerous.

To prevent the formation of peroxides, diethyl ether should be stored in airtight containers with added inhibitors, such as hydroquinone or butylated hydroxytoluene (BHT). It should also be used within a short time frame and stored away from heat sources, sparks, and flames. Regularly checking the peroxide levels in the ether and discarding it when peroxides are detected can help minimize the risks associated with this volatile, flammable, and explosive chemical.

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