which mineral property is the least useful for identifying minerals and why? which mineral property is the least useful for identifying minerals and why? density is the least useful property for identification because different specimens of the same type of mineral can vary in size. color is the least useful property for identification because it is the most obvious property. color is the least useful property for identification because the same mineral type can be found in several different colors as a result of impurities in the mineral. streak is the least useful property for identification because it does not work for minerals with metallic lusters. streak is the least useful property for identification because the same mineral type can be found with several different colors of streak caused by impurities in the mineral.

Answers

Answer 1

The least useful mineral property for identifying minerals is color. This is because the same mineral type can be found in several different colors as a result of impurities in the mineral. Color is not a reliable identifying feature for minerals, as it can vary considerably within a single mineral species.

Some minerals, such as quartz, can be found in a wide range of colors,, such as white, pink, purple, yellow, brown, and even black, making it difficult to use color alone to identify them. Mineral properties such as streak, hardness, cleavage, and luster provide more reliable information for mineral identification.

Density is also considered a less useful property for mineral identification because different specimens of the same type of mineral can vary in size, making it difficult to determine the density. Streak is also less useful for mineral identification because it does not work for minerals with metallic lusters, but it can be useful in identifying minerals that have non-metallic lusters.

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Related Questions

how many 30.0-g ice cubes are required to absorb 40.3 kj from a glass of water upon melting? the heat of fusion for water is 6.02 kj/mol.

Answers

Given: Heat absorbed by ice cubes Q = 40.3 KJ, Mass of one ice cube m = 30 g, Heat of fusion of water Hf = 6.02 KJ/mol

We need to find out how many 30.0-g ice cubes are required to absorb 40.3 kj from a glass of water upon melting. Therefore, we can solve this problem as follows: Firstly, let us find out the amount of heat required to melt one ice cube of 30 g.  
This is given by:
Q1 = m × HfQ1 = (30 g) / (1g/mol) × 6.02 kj/molQ1 = 180.6 kj/mol
Therefore, it takes 180.6 kJ to melt one ice cube of 30g
.Next, we can find out the number of ice cubes required as follows:
Number of ice cubes = Heat absorbed / Heat required to melt one ice cube= 40.3 kj / 180.6 kj/mol= 0.223 mol

Therefore, the answer is:0.223 × (6.02 × 10³ J/mol) / (30.0 g/mol)= 44.6 30.0-g ice cubes.

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For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH

A) 0.11 M NaOH

B) 1.5 x 10^-3 M Ca(OH)2

C) 4.8 x 10^-4 M Sr(OH)2

D) 8.7 x 10^-5 M KOH

Answers

A) 0.11 M NaOH:

Since NaOH is a strong base, it dissociates completely in water:

[OH-] = 0.11 M

[H₃O+] = 1 x10⁻¹⁴ / [OH-] = 1 x 10⁻¹⁴ 0.11 = 9.09 x 10⁻¹⁴M

pOH = -log[OH-] = -log(0.11) ≈ 0.96

pH = 14 - pOH = 14 - 0.96 ≈ 13.04

B) 1.5 x 10⁻³ M Ca(OH)₂:

Ca(OH)₂ dissociates to form two OH- ions per formula unit:

[OH-] = 2 x 1.5 x 10⁻³ = 3 x 10⁻³ M

[H3O+] = 1 x 10⁻¹⁴ / [OH-] = 1 x 10⁻¹⁴/ 3 x 10⁻³ = 3.33 x 10⁻¹²M

pOH = -log[OH-] = -log(3 x 10⁻³) ≈ 2.52

pH = 14 - pOH = 14 - 2.52 ≈ 11.48

C) 4.8 x 10⁻⁴ M Sr(OH)₂:

Sr(OH)₂ dissociates to form two OH- ions per formula unit:

[OH-] = 2 x 4.8 x 10⁻⁴ = 9.6 x 10⁻⁴ M

[H3O+] = 1 x 10⁻¹⁴/ [OH-] = 1 x 10⁻¹⁴ / 9.6 x 10⁻¹⁴ = 1.04 x 10⁻¹¹M

pOH = -log[OH-] = -log(9.6 x 10⁻⁴) ≈ 3.02

pH = 14 - pOH = 14 - 3.02 ≈ 10.98

D) 8.7 x 10⁻⁵ M KOH:

Since KOH is a strong base, it dissociates completely in water:

[OH-] = 8.7 x 10⁻⁵ M

[H3O+] = 1 x 10⁻¹⁴ / [OH-] = 1 x 10⁻¹⁴ / 8.7 x 10⁻⁵ = 1.15 x 10⁻¹⁰ M

pOH = -log[OH-] = -log(8.7 x 10⁻⁵) ≈ 4.06

pH = 14 - pOH = 14 - 4.06 ≈ 9.94

To determine the concentrations of hydroxide ions ([OH-]), hydronium ions ([H3O+]), pH, and pOH for the given strong base solutions, we can use the fact that strong bases dissociate completely in water. Here are the calculations for each solution:

A) 0.11 M NaOH:

Since NaOH is a strong base, it dissociates into Na+ and OH- ions. Therefore, [OH-] is equal to the concentration of NaOH, which is 0.11 M. In water, the concentration of H₃O+ is negligible because NaOH does not provide H+ ions. As a result, the pH can be calculated by taking the negative logarithm of the [OH-] concentration, which is approximately 13.04. The pOH is the negative logarithm of the [H₃O+] concentration, which is negligible.

B)  1.5 x 10⁻³ M Ca(OH)₂:

Calcium hydroxide ( Ca(OH)₂) dissociates into Ca₂+ and two OH- ions. Since the concentration of  Ca(OH)₂ is 1.5 x 10⁻³ M, the concentration of OH- ions is twice that, or 3 x 10⁻³ M. The concentration of H₃O+ is negligible in this case. Therefore, the pOH can be calculated by taking the negative logarithm of the [OH-] concentration, resulting in approximately 2.52. The pH is 14 minus the pOH, which is approximately 11.48.

C) 4.8 x 10⁻⁴ M Sr(OH)₂:

Strontium hydroxide (Sr(OH)2) dissociates into Sr₂+ and two OH- ions. Thus, the concentration of OH- ions is twice the concentration of Sr(OH)2, which is 9.6 x 10⁻⁴ M. Since the concentration of H₃O+ is negligible, the pOH can be calculated as approximately 3.02. The pH is 14 minus the pOH, which is approximately 10.98.

D)8.7 x 10⁻⁵ M KOH:

As KOH is a strong base, it dissociates into K+ and OH- ions. Consequently, the concentration of OH- ions is equal to the concentration of KOH, which is 8.7 x 10⁻⁵ M. Since there are no H₃O+ ions provided by KOH, the pH is calculated by taking the negative logarithm of the [OH-] concentration, resulting in approximately 9.94. The pOH is negligible in this case.

These calculations provide the values for [OH-], [H₃O+], pH, and pOH for each of the given strong base solutions.

how many carbonate ions are present in 30.0 mL of 0.600 M K2CO3 solution? With detail explain

Answers

There are approximately 2.17 × 10^22 carbonate ions in 30.0 mL of a 0.600 M K2CO3 solution.

The number of carbonate ions present in a solution of potassium carbonate (K2CO3), we need to use the concentration (Molarity) and volume of the solution.

Given:

Volume of K2CO3 solution = 30.0 mL = 0.0300 L (converted to liters)

Molarity of K2CO3 solution = 0.600 M

The molarity (M) of a solution represents the number of moles of solute per liter of solution. In this case, the solute is K2CO3, and the molarity is 0.600 M. This means that there are 0.600 moles of K2CO3 in every liter of solution.

The number of moles of K2CO3 in the given volume, we can use the formula:

moles = Molarity × Volume (in liters)

moles = 0.600 M × 0.0300 L

moles = 0.018 moles of K2CO3

Since K2CO3 contains two moles of carbonate ions[tex](CO3^2^-)[/tex]per mole of K2CO3, we can determine the number of moles of carbonate ions:

moles of [tex](CO3^2^-)[/tex]= 2 × moles of K2CO3

moles of [tex](CO3^2^-)[/tex]- = 2 × 0.018 moles

moles of [tex](CO3^2^-)[/tex] = 0.036 moles of CO3^2-

Finally, to convert moles to the number of particles (ions), we can use Avogadro's number (6.022 × 10^23 ions per mole):

number of carbonate ions = moles of CO3^2- × Avogadro's number

number of carbonate ions = 0.036 moles × 6.022 × 10^23 ions/mole

number of carbonate ions ≈ 2.17 × 10^22 carbonate ions

Therefore, there are approximately 2.17 × 10^22 carbonate ions in 30.0 mL of a 0.600 M K2CO3 solution.

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one model is composed to two atoms of the same element that is model...​

Answers

Answer:

nitrogen

Explanation:

NN pic at last

I need an answer quickly!

Answers

Titanium because it is

The weight of an object (in N) is equal to its mass
(in kg) multiplied by the acceleration of gravity (g),
which is equal to 9.8 m/s2 on Earth. What is the
weight of a person whose mass is 40 kg?


Help me plz.

Answers

Answer:

392 N

Explanation:

Weight=Mass*Gravity

Weight=40*9.8

Weight=392 Newtons

Amount (mol) of Cl atoms in 0.0374 g of C2H4Cl2

Answers

Answer:

98.96 g/mol..........

what volume of a 0.250 m naoh solution would be required to neutralize 155ml of a 0.225 m h3po4 solution?

Answers

Approximately 172.222 mL of a 0.250 M NaOH solution would be required to neutralize 155 mL of a 0.225 M H3PO4 solution.

To determine the volume of a 0.250 M NaOH solution required to neutralize 155 mL of a 0.225 M H3PO4 solution, we need to use the concept of stoichiometry and the balanced chemical equation between NaOH and H3PO4.

The balanced equation between NaOH and H3PO4 is:

3NaOH + H3PO4 -> Na3PO4 + 3H2O

From the balanced equation, we can see that it takes three moles of NaOH to neutralize one mole of H3PO4. This means that the mole ratio between NaOH and H3PO4 is 3:1.

To calculate the volume of NaOH solution needed, we need to use the molarity and volume of the H3PO4 solution and set up a proportion based on the stoichiometry:

0.225 moles H3PO4 / 155 mL H3PO4 = 0.250 moles NaOH / x mL NaOH

Cross-multiplying and solving for x:

x = (155 mL H3PO4 * 0.250 moles NaOH) / 0.225 moles H3PO4

x = 172.222 mL NaOH

Using the stoichiometry and the balanced equation between NaOH and H3PO4, we find that approximately 172.222 mL of a 0.250 M NaOH solution is needed to neutralize 155 mL of a 0.225 M H3PO4 solution. The calculation involves setting up a proportion based on the mole ratio between NaOH and H3PO4.

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a marble with a mass of 14.1 grams is placed inside a graduated cylinder. the volume of water in the graduated cylinder rises from 10.0 ml to 13.7 ml. calculate the density of the marble.

Answers

To find the density of a marble, divide its mass by its volume. The volume of the marble can be calculated by subtracting the initial volume (10.0 ml) from the final volume (13.7 ml).

Thus, the volume of the marble is:13.7 ml - 10.0 ml = 3.7 mlTo convert this to cubic centimeters (cc), we need to multiply by a conversion factor of 1 cc/1 ml. Thus, the volume of the marble in cc is:3.7 ml × (1 cc/1 ml) = 3.7 cc

Now, we can find the density of the marble by dividing its mass (14.1 grams) by its volume (3.7 cc):

Density = Mass/VolumeDensity

             = 14.1 g/3.7 ccDensity

             ≈ 3.81 g/cc

Therefore, the density of the marble is approximately 3.81 g/cc.

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What is the function of an Adams nucleus

Answers

Answer:

The key function of the nucleus is to control cell growth and multiplication. This involves regulating gene expression, initiating cellular reproduction, and storing genetic material necessary for all of these tasks. In order for a nucleus to carry out important reproductive roles and other cell activities, it needs proteins and ribosomes.

Explanation:

Answer:

sorry need to awnser 2 question dont have musch more time

Explanation:

whats an atomic nucleus for middle school
here are the choses
A:the outer part of an atom in which electrons move
B:the outer part of an atom in which neutrons move
C:the central part of an atom, composed of protons and neutrons
D:the central part of an atom, composed of protons and electrons

Answers

Answer:

C

Explanation:

The nucleus contains protons and neutrons, electrons are located in outer electron cloud.

Answer:

C is the answer

Explanation:

hoped this helped

The temperature of a substance is 329 K. What is the
temperature in degrees Celsius (°C)?


Hi
How you guys doing?
Help me plz.

Answers

Answer: 55.85 c

Explanation:

Answer:

55.85°C

Explanation:

Temp Conversion: K - 273.15 = °C

Step 1: Define variables

K = 329 K

°C = ?

Step 2: Substitute and Evaluate for °C

329K - 273.15 = °C

55.85°C

How does heat transfer from the inner core from the asthenosphere?

Answers

Answer:

Heat is transferred to the surface of the Earth from the hot Earth's core by conduction and from radiation from the Sun. The remaining heat on the surface is sent out into space in the form of infrared radiation.

hope this help you.

Why couldn't you substitute concentrated HNO_3 solution for 3M H_2 SO_4 in Parts IV and V? Neither acid should be used If HNO_3 solution were used in Parts IV and V, Cu(OH)_2 would not be reduced to Cu by Zn, because HNO_3 solution oxidizes Cu to Cu(OH)_2 Both acids could be used If HNO_3 solution were used in Parts IV and V, [Cu(H_2 O)_6]^2+ ion would not be reduced to Cu by Zn, because HNO_3 solution oxidizes Cu to [Cu(H_2 O)_6]^2+

Answers

Concentrated HNO3 solution cannot be substituted for 3M H2SO4 in Parts IV and V because HNO3 solution oxidizes copper (Cu) to Cu(OH)2 and [Cu(H2O)6]^2+, preventing the reduction of Cu(OH)2 to Cu by Zn. Therefore, H2SO4 is the preferred acid for these parts of the experiment.

In Parts IV and V of the experiment, the reduction of Cu(OH)2 to Cu is carried out using Zn as the reducing agent. To facilitate this reduction reaction, it is crucial to use an acid that does not oxidize copper.

Concentrated HNO3 solution is not suitable for these parts because it oxidizes copper. It would react with Cu, converting it to Cu(OH)2 or even further to the complex ion [Cu(H2O)6]^2+.

On the other hand, 3M H2SO4 is a suitable acid for these parts because it does not oxidize copper. It provides the necessary acidic conditions for the reduction reaction to occur without interfering with the copper species.

Therefore, to ensure the successful reduction of Cu(OH)2 to Cu by Zn, it is important to use 3M H2SO4 as specified in Parts IV and V of the experiment.

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which tool would best help, will mark brainlist to first 1

Answers

Answer:

c

Explanation:

C. hand lens would magnify the object.

Use data from Appendix IIB to calculate the equilibrium constants at 25 ∘C∘C for each of the following reactions.
Part A
2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g)
K = ?
Express your answer using two significant figures.
Part B
Br2(g)+Cl2(g)⇌2BrCl(g)Br2(g)+Cl2(g)⇌2BrCl(g)
ΔG∘fΔGf∘ for BrCl(g)BrCl(g) is -1.0 kJ/molkJ/mol
K = ?
Express your answer using two significant figures.

Answers

A. Due to the unavailability of Appendix IIB data, it is not possible to calculate the equilibrium constant K for the given reaction at 25°C.

B. The equilibrium constant, K, for the reaction Br2(g) + Cl2(g) ⇌ 2BrCl(g) at 25°C is approximately 3.34.

Part A

The equilibrium constant, K, is given by the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

In this case, the reaction is:

2NO2(g) ⇌ N2O4(g)

The equilibrium constant expression is:

K = [N2O4] / [NO2]²

To calculate K at 25°C, we need to refer to Appendix IIB, which provides the standard Gibbs free energy change (ΔG°) values for the reaction at different temperatures.

Unfortunately, without Appendix IIB data, it is not possible to calculate the equilibrium constant K at 25°C.

Due to the unavailability of Appendix IIB data, it is not possible to calculate the equilibrium constant K for the given reaction at 25°C.

Part B:

The equilibrium constant, K, can be determined using the relationship between ΔG° and K. The standard Gibbs free energy change, ΔG°, is related to K through the equation:

ΔG° = -RT ln(K)

where R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln denotes the natural logarithm.

Given that ΔG°f for BrCl(g) is -1.0 kJ/mol, we can convert it to joules by multiplying by 1000:

ΔG°f = -1.0 kJ/mol

ΔG°f = -1000 J/mol

Now, let's substitute the values into the equation:

-1000 J/mol = -8.314 J/(mol·K) × 298 K × ln(K)

Simplifying the equation:

ln(K) = -1000 J/mol / (-8.314 J/(mol·K) × 298 K)

ln(K) = 1.205

To find K, we exponentiate both sides:

K = e^1.205

Using a calculator, we find:

K ≈ 3.339

The equilibrium constant, K, for the reaction

Br2(g) + Cl2(g) ⇌ 2BrCl(g) at 25°C is approximately 3.34.

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Susan conducts an experiment five times and gets a solution concentration of 1.9M, 2.1M, 1.8M, 1.9M, and 2.2M. The known concentration of the solution is 2.0M. Which of the following are true about Susan's results?



A. They are both accurate and precise

B. They are precise but not accurate

C. They are accurate but not precise

D. They are neither accurate nor precise

Answers

d they are accurate but not preci

A calorimeter was used to determine the molar enthalpy of a solution of AICI3. When a sample of AICI3, was dissolved in water, the following data were collected.
Mass of AlCl3 (g) : 10.0
Mass of H2O (g) : 250.0
Initial Temp. (°C) : 20.0
Final Temp. (°C) : 46.3
Determine the molar heat (kJ/mol) of solution. Answer in 3 Sig fig, include sign.

Answers

The molar heat of solution of AICI3 is approximately -120 kJ/mol, indicating that the dissolution of AICI3 in water is an exothermic process.

To determine the molar heat of solution, we can use the formula:

q = m * C * ΔT

where:

q is the heat absorbed or released by the solution,

m is the mass of the solution in grams,

C is the specific heat capacity of the solution (assumed to be the same as water, 4.18 J/g°C),

ΔT is the change in temperature.

First, let's convert the given mass of AlCl3 and H2O to moles:

- Mass of AlCl3: 10.0 g

- Molar mass of AlCl3: 133.34 g/mol (Al: 26.98 g/mol, Cl: 35.45 g/mol)

- Moles of AlCl3: 10.0 g / 133.34 g/mol = 0.075 mol

- Mass of H2O: 250.0 g

- Molar mass of H2O: 18.02 g/mol (H: 1.01 g/mol, O: 16.00 g/mol)

- Moles of H2O: 250.0 g / 18.02 g/mol = 13.87 mol

Next, let's calculate the heat absorbed or released by the solution:

q = m * C * ΔT

Since the specific heat capacity is given in J/g°C, we need to convert the units to kJ/mol°C:

1 J/g°C = 0.239 kJ/mol°C

q = (10.0 g + 250.0 g) * 0.239 kJ/mol°C * (46.3°C - 20.0°C)

q = 260.0 g * 0.239 kJ/mol°C * 26.3°C

q = 1627.22 kJ

Finally, let's calculate the molar heat of solution:

Molar heat of solution = q / (moles of AlCl3 + moles of H2O)

Molar heat of solution = 1627.22 kJ / (0.075 mol + 13.87 mol)

Molar heat of solution ≈ 120.43 kJ/mol

Rounded to three significant figures, the molar heat of solution of AICI3 is approximately -120 kJ/mol.

The molar heat of solution of AICI3 is approximately -120 kJ/mol, indicating that the dissolution of AICI3 in water is an exothermic process.

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Suppose the interaction between two atoms by the Lennard-Jones potential: ULJ = B/r^12 - A / r^6 where the values of A and B are known to be A = 10^-77 J x m^6 and B = 10^-134 J x m^12.
What does the Lennard-Jones potential predict for the separation r=r eq
​hen the energy is at the minimum (equilibrium) value, U min. What is the u min fot this interaction at T=298 K ? What is the ratio of U min to the purely attractive van der Waals component of the interaction potential at r eq.
What is the ratio of r eq to r 0 defined by u(r 0 )=0. 4. What is the ratio of r s to r 0 , where r s is the separation where the magnitude of the (attractive adhesion) force is maximum, F max . What is the value for F max between the two atoms?

Answers

a) The Lennard-Jones potential predicts the separation r_eq at the minimum energy U_min.

b) The U_min for this interaction at T=298 K is the value obtained from the Lennard-Jones potential equation when r=r_eq.

c) The ratio of U_min to the purely attractive van der Waals component of the interaction potential at r_eq can be calculated by comparing the attractive part (-A/r^6) to the total potential energy U_min.

d) The ratio of r_eq to r_0, where u(r_0)=0.4, can be determined by finding the value of r_eq where the potential energy is equal to 0.4 times the total potential energy at r=r_0.

e) The ratio of r_s to r_0, where r_s is the separation where the magnitude of the attractive adhesion force is maximum, can be determined by finding the value of r where the derivative of the potential energy with respect to r is equal to zero.

f) The value of F_max between the two atoms can be obtained by taking the negative derivative of the potential energy equation with respect to r and evaluating it at r=r_s.

a) The Lennard-Jones potential provides information about the relationship between energy and separation between two interacting atoms.

At the minimum energy (U_min), the potential predicts the separation r_eq, which corresponds to the equilibrium distance between the atoms. This is the distance at which the energy of the system is at its lowest point.

b) To determine the value of U_min at a given temperature (T=298 K), you can substitute the equilibrium separation r_eq into the Lennard-Jones potential equation and calculate the resulting energy value.

This will give you the U_min for the interaction.

c) The Lennard-Jones potential consists of two components: an attractive component (-A/r^6) and a repulsive component (B/r^12).

The ratio of U_min to the purely attractive van der Waals component of the interaction potential at r_eq can be calculated by comparing the magnitude of the attractive component to the total potential energy at the equilibrium separation.

This ratio provides insights into the relative contribution of the attractive force to the overall potential energy at equilibrium.

d) The ratio of r_eq to r_0 can be determined by finding the value of r_eq where the potential energy is equal to 0.4 times the total potential energy at r=r_0.

In other words, you need to solve the Lennard-Jones potential equation for r_eq when the potential energy is equal to 0.4 times the potential energy at r=r_0.

e) The ratio of r_s to r_0 is obtained by finding the value of r where the magnitude of the attractive adhesion force is maximum.

This can be determined by finding the separation r where the derivative of the potential energy equation with respect to r is equal to zero.

The value of r_s represents the separation at which the attractive force between the atoms is strongest.

f) The value of F_max between the two atoms can be obtained by taking the negative derivative of the Lennard-Jones potential energy equation with respect to r and evaluating it at r=r_s.

This will give you the magnitude of the maximum attractive adhesion force between the atoms.

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What does the crossing over rule essentially help us figure out? In chemistry

Answers

Either the charge of the element, or the number of atoms present in that element

Select all that apply Which of the following statements correctly describe the VSEPR model? Select all that apply. a. According to this model, the valence electrons around a central atom are located as far from each other as possible b. A Lewis structure is necessary in order to apply the VSEPR model.
c. The VSEPR model states that the electron domains surrounding a central atom attract each other. d. The VSEPR model is used to predict the number of bonds a given element will form. e. The VSEPR model is used to predict the geometry of a covalently bonded species

Answers

The correct statements are: a. According to the VSEPR model, the valence electrons around a central atom are located as far from each other as possible,

e. The VSEPR model is used to predict the geometry of a covalently bonded species.

These two statements correctly describe the VSEPR model. The VSEPR model is based on the principle that electron pairs (both bonding and nonbonding) around a central atom repel each other, leading to the arrangement of these electron pairs in a way that maximizes their separation.

This results in the prediction of the molecular geometry or shape of a covalently bonded species.

The other statements (b, c, d) are incorrect descriptions of the VSEPR model. A Lewis structure is not necessary to apply the VSEPR model, as it focuses on the arrangement of electron pairs rather than the specific bond formation.

The VSEPR model does not state that electron domains surrounding a central atom attract each other, but rather that they repel each other. Additionally, the VSEPR model is not primarily used to predict the number of bonds a given element will form, but rather the spatial arrangement of electron pairs.

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In your own words what is Substance. In your own words what is Mixture

Answers

Answer:

Subtance is a matter such as gold,copper, and oxygen.

Mixture is when you combine two different substances.

Mixtures are one product of mechanically blending.

Oxalic acid (98%) is a polyprotic acid. It has a density of 1.65 g/cm^3 and a melting point of 189.5°C. Oxalic acid has a molecular mass of 90.03 g/mol and with a pka1 of 5.62 x10^-2. What volume of oxalic acid must be added to sufficient water to give a 1.500 liter solution that is 0.300 F (in formal concentration)?

Answers

Approximately 24.55 cm^3 of oxalic acid must be added to sufficient water to give a 1.500 liter solution with a formal concentration of 0.300 F.

To find the volume of oxalic acid needed to make a 1.500 liter solution with a formal concentration of 0.300 F, we need to use the equation:

Formal concentration (F) = (moles of solute) / (volume of solution in liters)

First, we need to calculate the moles of oxalic acid required. The formal concentration (F) is given as 0.300, so:

0.300 = (moles of oxalic acid) / 1.500

Rearranging the equation, we find:

moles of oxalic acid = 0.300 * 1.500

moles of oxalic acid = 0.450

Next, we can calculate the mass of oxalic acid needed using its molecular mass:

mass of oxalic acid = moles of oxalic acid * molecular mass

mass of oxalic acid = 0.450 * 90.03

mass of oxalic acid = 40.5145 g

Finally, we can calculate the volume of oxalic acid needed using its density:

volume of oxalic acid = mass of oxalic acid / density

volume of oxalic acid = 40.5145 g / 1.65 g/cm^3

volume of oxalic acid = 24.55 cm^3

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.Which element has atoms in the ground state with the greatest number of valence electrons?
A)tin
B)sulfur
C)arsenic
D)fluorine

Answers

D) fluorine  has atoms in the ground state with the greatest number of valence electrons.

Among the options given, fluorine (F) has atoms in the ground state with the greatest number of valence electrons. Valence electrons are the electrons in the outermost energy level (also known as the valence shell) of an atom. The number of valence electrons determines the chemical behavior and reactivity of an element.

Fluorine is located in Group 17 (Group VIIA) of the periodic table, also known as the halogens. It has 9 electrons in its outermost shell, corresponding to its atomic number of 9. This means that fluorine has a full 2s orbital and 7 electrons in the 2p orbital, resulting in a total of 7 valence electrons.

In comparison, tin (Sn) has 4 valence electrons, sulfur (S) has 6 valence electrons, and arsenic (As) has 5 valence electrons.

Therefore, among the given options, fluorine has the greatest number of valence electrons in its ground state.

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How much energy will a photon with a frequency of 6.8 x 104 Hz emit?

Answers

Answer:

Energy emitted by photon = 45.06 × 10−32 J

Explanation:

Energy of photon is given by  E = hf

where

h is the Planck's constant whose value is 6.626×10−34J⋅s

f is the frequency of photon

________________________________________

given that

f = 6.8 x 104 Hz

1 HZ = 1 s^-1

E =  6.626×10−34J⋅s * 6.8 x 10^4 Hz

E = 45.06 × 10−32 J

Thus,

Energy emitted by photon is 45.06 × 10−32 J

Which of the following is an ionic compound?
O CO2
O H2O
O CH4
OKCI

Answers

Answer:

the first one since it mean carbon oxide

Explanation:

and u need a cation and anion to create a compound which is what that is

A buffer contains significant amounts of acetic acid and sodium acetate.
a)Write an equation showing how this buffer neutralizes added acid (HCl).
b)Write an equation showing how this buffer neutralizes added base (NaOH).
**Write as an chemical equation**

Answers

a) The equation for how the buffer neutralizes added acid (HCl) is: CH3COOH + HCl -> CH3COOH2+ + Cl-

b) The equation for how the buffer neutralizes added base (NaOH) is: CH3COOH + OH- -> CH3COO- + H2O

a) When an acid, such as HCl, is added to the buffer, the acetic acid (CH3COOH) in the buffer reacts with the added acid. The equation for this reaction is CH3COOH + HCl -> CH3COOH2+ + Cl-. The acetic acid donates a proton (H+) to the HCl, forming the hydronium ion (CH3COOH2+) and the chloride ion (Cl-). This reaction helps to neutralize the added acid and maintain the pH of the buffer solution.

b) When a base, such as NaOH, is added to the buffer, the acetic acid in the buffer reacts with the added base. The equation for this reaction is CH3COOH + OH- -> CH3COO- + H2O. The hydroxide ion (OH-) from the base reacts with the acetic acid, resulting in the formation of the acetate ion (CH3COO-) and water (H2O). This reaction helps to neutralize the added base and maintain the pH of the buffer solution.

In both cases, the presence of both acetic acid and its conjugate base, sodium acetate, in the buffer system allows for the buffering capacity of the solution. The acetic acid can act as a proton donor (acid), while the acetate ion can act as a proton acceptor (base), helping to maintain the pH of the solution within a specific range.

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if you have 500 ml of a 0.10 m solution of the acid, what mass of the corresponding sodium salt of the conjugate base do you need to make the buffer with a ph of 3.48 (assuming no change in volume)?

Answers

To determine the mass of the corresponding sodium salt of the conjugate base that is required to make the buffer with a pH of 3.48, we need to use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is as follows; pH = pKa + log ([A-]/[HA]), where [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the acid. pH = 3.48, pKa is the negative log of the acid dissociation constant, and it is given as 4.75, and [A-]/[HA] can be calculated from the initial concentration of the acid solution;[A-]/[HA] = 10^(pH - pKa) = 10^(3.48 - 4.75) = 0.0629M. Next, we can use the concentration of the conjugate base and the volume of the buffer to determine the number of moles of the conjugate base that we need.0.0629 M x 0.5 L = 0.03145 molWe can then use the molar mass of the sodium salt of the conjugate base to calculate the mass required. This can be obtained from the molecular formula of the salt. The molecular formula of the sodium salt of the conjugate base is not given in the question. Hence, we cannot determine the molar mass and the mass required. To make a buffer of a specific pH, we require an acid and its corresponding conjugate base. The pH of a buffer can be calculated using the Henderson-Hasselbalch equation.

To make the buffer of pH 3.48, we require the corresponding sodium salt of the conjugate base and the acid. Using the Henderson-Hasselbalch equation, we determined the concentration of the conjugate base required. We then used the volume of the buffer to determine the number of moles of the conjugate base required. However, we could not determine the molar mass of the sodium salt of the conjugate base and hence could not determine the mass required.

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Atoms of two different elements must have different
1. electrical charges.
2. numbers of neutrons.
3. atomic numbers
4. energy levels.​

Answers

Answer:

C atomic number

Explanation:

As u go from left to right the atomic number changes

So C is the answer,

I took the test on edge and that was the answer

Explain what hypotonic, hypertonic and isotonic mean in terms of osmolarity

Answers

Hypotonic, hypertonic, and isotonic are terms used to describe the concentration of solutes in a solution compared to the concentration of solutes in a cell or another solution.

Hypotonic solutions have a lower osmolarity (concentration of solutes) compared to the inside of a cell. When a cell is placed in a hypotonic solution, water will flow into the cell in an attempt to equalize the concentrations. This causes the cell to swell and potentially burst. Hypotonic solutions are commonly used in medical settings to hydrate patients and replenish fluid loss.

Hypertonic solutions have a higher osmolarity compared to the inside of a cell. When a cell is placed in a hypertonic solution, water will flow out of the cell, causing it to shrink. Hypertonic solutions are often used to draw excess fluid out of swollen tissues or to dehydrate and preserve food.

Isotonic solutions have the same osmolarity as the inside of a cell. When a cell is placed in an isotonic solution, there is no net movement of water. Isotonic solutions are frequently used in medical treatments, such as intravenous fluids, to maintain proper fluid balance and prevent cell damage.

In summary, hypotonic solutions have a lower solute concentration, hypertonic solutions have a higher solute concentration, and isotonic solutions have an equal solute concentration compared to the inside of a cell.

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