Which of these is true for the electrons of an atom? Choose

What is the source of fuel that is most widely used today?

34kurt

The <span>source of fuel that is most widely used today is Natural gas. The answer is letter C. The rest of the choices do not answer the question above,</span>

8 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

what volume of 1.75 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3 ?

Evgesh-ka [11]

The balance chemical equation is :

2HCL + Na2CO3 -----> 2NaCl + H2O + CO2

According to Question,

Given,

Molarity of HCL = 1.75 m

Sodium Carbonate (Na2CO3) = 0.100 m

For finding out the molarity,

c = n solute / Vsolution => n = c.Vsolution

nCO^2-3 = 0.100 mol L^-1 . 0.750 L = 0.0750 moles of CO^2-3

0.0750 moles of CO^2-3 . 2 moles H3O^+ / 1 mole CO^2-3

= 0.150 moles of H3O^+

As we have already know the molarity of HCL , we easily calculate what volumes by many moles.

c = nsolute / Vsolution => Vsolution = nsolute / c

VH3O^+ = 0.150 moles / 1.75 mol L^-1 = 0.0857 L

To know more about Molarity here :

brainly.com/question/8732513?referrer=searchResults

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6 0

1 year ago

A solution of acetic acid, CH3CO₂H(aq), is at equilibrium. How would the

s344n2d4d5 [400]

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation. (Option C)

<h3>How do systems at equilibrium respond to perturbation?</h3>

When a system at equilibrium suffers a perturbation, it shifts its equilibrium position to counteract such perturbation.

Let's consider a solution of acetic acid at equilibrium.

CH₃CO₂H(aq) = CH₃CO₂⁻(aq) + H⁺(aq)

If more acetic acid were added to the solution, the system will shift toward the products to counteract such an increase.

How would the system change if more acetic acid were added to the solution?

A. [H⁺] would decrease and [CH₃CO₂⁻] would increase. NO.

B. [H⁺] and [CH₃CO₂⁻] would decrease. NO.

C. [H⁺] and [CH₃CO₂⁻] would increase. YES. Both products would increase.

D. [H⁺] would increase and [CH₃CO₂⁻] would decrease. NO.

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation.

Learn more about equilibrium here: brainly.com/question/2943338

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3 0

2 years ago

Three test tubes contain white crystalline organic solids A, B, and C, each of which melts at 149-150 degrees C. A 50-50 mixture

Digiron [165]

Answer:

The mixture of B and C will melt as 130 - 139 ºC.

Explanation:

If the melting point (130 - 139 ºC) of a mixture of A and B is lower than the pure substances that is 149 - 150 ºC that means that one of these susbtances is an impurity because reduces the melting point.

If the melting point of the mixture of A and C is the same as the pure substances, we can deduce that they are the same substances, also because the melting point when a substance is pure is just 1 or 2 ºC like in this case.

So in a Mixture of B and C is going to be like the first case of the mixture among A and B, because the B substance is the impurity.

3 0

3 years ago