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kifflom [539]
3 years ago
9

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

)]2+ that would be formed from 6.00 mL of 0.00180 M KSCN 5.00 mL 0.240 M Fe(NO3)3 and 14.00 mL of 0.050 M HNO3.
Question 3 options:

0.240 M

4.32 x 10^-4

0.0480

0.0460 M
Chemistry
1 answer:
viktelen [127]3 years ago
7 0

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}

\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol

Moles of KSCN = Moles of K^+ = Moles of SCN^- = 1.08\times 10^{-5}mol

Now we have to calculate the concentration of [Fe(SCN)]^{2+}

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M

Thus, the concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

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Answer:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

  • an equilibrium constant is, first of all, a fraction;
  • in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
  • in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
  • each concentration should be raised to the power of the coefficient in the balanced chemical equation;
  • only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}

4 0
3 years ago
Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
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Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

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<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

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Is the density of iron greater then the density of gold
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Gold is one of the softest metals. So YES

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Consider the reaction: 2 H2O (g)--&gt;2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag
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DE = dH - PdV 

<span>2 H2O(g) → 2 H2(g) + O2(g) </span>

<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>

<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>

<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>

<span>2 moles reacts to form 3 moles </span>

<span>The gas equation is </span>

<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
<span>R = gas constant = 8.314 J K^-1 mol^-1 </span>
<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>

<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>

<span>dE = dH -dnRT (because PV = nRT) </span>

<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>

<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>

<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>
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3 years ago
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Answer:

The system shifts forward

Explanation:

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