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3 years ago

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An industrial chemist introduces 8.1 atm h2 and 8.1 atm co2 into a 1.00-l container at 25.0°c and then raises the temperature to

700.0°c, at which keq = 0.534: h2(g) + co2(g) ⇔ h2o(g) + co(g) how many grams of h2 are present after equilibrium is established?

1 answer:

kirill115 [55]3 years ago

3 0

I reccomend searching it up

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What is true about empathy?

uysha [10]

I would go with B. Not everyone can learn this skill.

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3 years ago

A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0

yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml = 150 ml

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.178\times 1000}{150}=1.19M$

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = molarity of stock solution = 1.19 M

$V_1$ = volume of stock solution = 15.0 ml

$C_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 65.0 ml

Putting in the values we get:

$1.19\times 15.0=M_2\times 65.0$

$M_1=0.275M$

Therefore, the concentration of KOH for the final solution is 0.275 M

5 0

2 years ago

A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33

Burka [1]

Answer:

B. A precipitate will form since Q > Ksp for calcium oxalate

Explanation:

Ksp of CaC₂O₄ is:

CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻

Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹

In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

Replacing in Ksp formula:

[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.

Thus, right answer is:

<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>

<em></em>

4 0

3 years ago

What was the scientific basis for the unit amu

Naya [18.7K]

Natural abundance of oxygen I think

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3 years ago

How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant

schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

1 g = 1,000 mg
The molar mass of Cu is 63.55 g/mol
When 1 mole of Cu is deposited, 2 moles of electrons circulate.
The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
1 A = 1 C/s

The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

$336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s$

3 0

3 years ago