Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T[tex]_2[/tex] ),

P[tex]_1[/tex]V[tex]_1[/tex] / T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / T[tex]_2[/tex],

T[tex]_2[/tex] = P[tex]_2[/tex]V[tex]_2[/tex]T[tex]_1[/tex] / P[tex]_1[/tex]V[tex]_1[/tex],

T[tex]_2[/tex] = 0.73 atm [tex]*[/tex] 1.8 L [tex]*[/tex] 298.15 K / 1 atm [tex]*[/tex] 1.2 L = ( 0.73 [tex]*[/tex] 1.8 [tex]*[/tex] 298.15 / 1 [tex]*[/tex] 1.2 ) K = 326.47425 K,

T[tex]_2[/tex] = 326.47425 K = 53.32425 C

Answer:

211.47 grams

Explanation:

We need to set up a dimensional analysis to solve this problem by converting from moles to grams.

First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:

1.01 + 35.45 = 36.46 g/mol

We have 5.8 moles of HCl, so multiply by its molar mass:

(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g

The answer is thus 211.47 grams.

~ an aesthetics over

Answer:

[tex]\large\boxed{211.47}\\[/tex] grams

Explanation:

First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.

[tex]\large\boxed{211.47}[/tex] is the final answer.

Answer:

1, C, Xe 2, B,He

Explanation:

1, cause as u go down a group the boiling point increases.

2, boiling point of single element is greater than a compound

According to  periodic trends in periodic table boiling point increases down the  group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.

Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.

It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.

Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.

Learn more about periodic table,here:

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Answer:

e. reducing the pressure from 608 torr to 0.40 atm at constant temperature.

Explanation:

According to Boyle's law when a gas is at the same temperature and there is a mass in a closed container so the pressure and the volume changes in the opposite direction

So here the equation is

[tex]P_1V_1=P_2V_2[/tex]

Now we choose the options

where,

[tex]V_1 = 100\ mL = 0.1\ L\\\\V_2 = 200\ mL = 0.2\ L[/tex]

[tex]P_1 = 608\ torr = 0.8\ atm \\\\P_2= 0.4\ atm[/tex]

Now applying these values to the above equation

So,

P1V1=P2V2

[tex]P_1V_1=P_2V_2[/tex]

[tex]0.8\times0.1 = 0.4\times0.2[/tex]

0.8 = 0.8

Hence, it is proved

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

Answer:

AgBr

Explanation:

Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.

However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.

The equation for the dissolution of AgBr in cyanide is shown below;

AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)

Answer:

The number of electrons transferred in the reaction

Explanation:

Answer:

A

Explanation:

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

Find more about activation energy here:

I hope it helps you!

Answer:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Explanation:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Answer: A=1, B=3, C=12

Explanation:

For this problem, you will need to know your unit conversions. There are 3 ft in 1 yard. Knowing this, we can find A, B, C.

For A and B, we know that we want to cancel out ft so the answer can be in yards. To do so, we need to put B=3 and A=1.

Now that we know the unit conversion, we can directly solve.

36 ft×(1 yd/3 ft)=12 yd

Our final answer is A=1, B=3, C=12.

Answer:

A=1,000, B=1, C=5,400

Explanation:

the question was 5.4L x AmL / BmL = CmL

Answer: C

Explanation:

The article was sourced from the Oak National Laboratory

Which reasons did you include in your response? Check all of the boxes that apply.

1. The article does not present original research.

and

3. The article has references to primary sources.

Answer:

C

Explanation:

Which reasons did you include in your response? Check all of the boxes that apply.

The article does not present original research.

The article summarizes other research.

The article has references to primary sources.

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E =4.057 \times 10^{-19} \ J[/tex]

Converting Joules (J) to eV ; we get,

[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

Answer:

(a )people separate mixtures in order to ger a specific substance that they need.

Answer:

Explanation:

Given:

Mass of solute = 30 gram

Volume of water = 500 ml

Density = 1.15g/ml

Find:

(i)ppm

(ii) %w/v

(iii) %w/w

Computation:

Water in gram = 500 ml × 1.15 g/ml

Water in gram = 575 gram

In ppm

Concentration of MgSO4 = [30 / 575] × 10⁶

Concentration of MgSO4 = 0.0521 × 10⁶ ppm

in % w/v

Concentration of MgSO4 = [30 / 500] × 100

Concentration of MgSO4 = 6% w/v

in % w/w

Concentration of MgSO4 = [30 / 575] × 100

Concentration of MgSO4 = 5.21% w/w

Answer:

310 mL

Explanation:

Step 1: Given data

Step 2: Calculate the initial volume

We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.400 M × 5.50 L / 7.10 M

V₁ = 0.310 L = 310 mL

Answer:

155 °C

Explanation:

Step 1: Given data

Step 2: Convert the initial temperature to Kelvin

We will use the following expression.

K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K

Step 3: Calculate the final temperature

Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 4.20 L × 296.2 K/2.91 L

T₂ = 428 K

Step 4: Convert the final temperature to Celsius

We will use the following expression.

°C = K - 273.15 = 428 - 273.15 = 155 °C

Answer:

[tex]x_{et}=0.6068[/tex]

Explanation:

Hello,

In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:

[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]

Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)

[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]

Therefore, the mole fraction turns out:

[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]

Best regards.

Answer:

Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.

Explanation:

Answer:

50.

Explanation:

We can write Graham's Law of Diffusion as:

(Rate 1)^2 = Molecular Mass 2

--------------    -------------------------

(Rate 2)^2    Molecular Mass 1

So using the Given Information:

1^2  / (1/5)^2 = Molecular Mass of unknown gas / 2, so:

25 = M/2

M = 50.

Answer:

A

Explanation:

The lithosphere represents the layer of hardened/solid rock that makes up the hard part of the earth, including the brittle upper portion of the mantle and the crust. The lithosphere is broken into pieces that are referred to as plates. The pieces move to and away from each other in a process known as plate tectonics. The movement of plates accounts for the global locations of volcanoes, earthquakes, and mountain ranges.

The lithosphere is made of largely of inorganic materials known as silicates. The weathering of the solid rocks together with the interaction of living organisms gives rise to soil with an appreciable amount of organic materials.

The correct option is, therefore A.

Answer:

[tex]Ksp=2.59x10^{-3}[/tex]

Explanation:

Hello,

In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:

[tex]AB_3\rightleftharpoons A^{3+}+3B^-[/tex]

The concentrations of the A and B ions in the solution are:

[tex][A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M[/tex]

[tex][B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M[/tex]

Then, as the solubility product is defined as:

[tex]Ksp=[A][B]^3[/tex]

Due to the given dissociation, it turns out:

[tex]Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}[/tex]

Regards.

Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

Answer:

2HClO(aq) + Ba(OH)₂(aq) →  Ba(ClO)₂(aq) + 2H₂O(l)

Explanation:

The reaction corresponds to a neutralization reaction between an acid and a base, as follows:

2HClO(aq) + Ba(OH)₂(aq)  →  Ba(ClO)₂(aq) + 2H₂O(l)            

From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.

In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.

I hope it helps you!    

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Answer:

We can then determine that a reaction will shift to the right if Q<K

Explanation:

Comparing Q with K allows to find out the status and evolution of the system:

In this case, we can then determine that a reaction will shift to the right if Q<K

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

Answer:

37.8

Explanation:

Answer:

Pentan-2-ol

Explanation:

On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ([tex]H^+[/tex])  would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".

See figure 1

I hope it helps!

Answer:

14.297 g

Explanation:

From the question;

1 mo of the compound requires 1320.0 kJ

From the molar mass;

1 ml of the compound weighs 30.55g

How many grams requires 617.30kJ?

1 ml = 1320

x mol = 617.30

x = 617.30 / 1320

x = 0.468 mol

But 1 mol = 30.55

0.468 mol = x

x = 14.297 g

Answer:

atoms contain tiny negatively charged subatomic particles or electrons

Answer:

An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction