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3 years ago

I’m just checking my answer. Thanks!

Chemistry

1 answer:

OLEGan [10]3 years ago

5 0

Answer:

78.89 g of CO is required.

Explanation:

$Fe_{2}O_{3}+3CO\longrightarrow3CO_{2}+2Fe$

Fe = 55.85 g/mol

O = 15.99 g/mol

C = 12.01 g/mol

2 moles of Fe is produced on reacting 1 mole of $Fe_{2}O_{3}$ .

2 moles of Fe is produced on reacting 3 moles of CO.

$Weight\:of\:one\:mole\:of\:Fe_{2}O_{3}=2\times55.85+3\times15.99=159.67g\\$

$Weight\:of\:one\:mole\:of\:CO=12.01+15.99=28g$

$The\:weight\:of\:two\:moles\:of\:Fe=2\times55.85=111.7g\\111.7g\:of\:Fe\:is\:produced\:by\:reacting\:(3\times28=)84g\:of\:CO\\\\104.9g\:of\:Fe\:is\:produced\:on\:reacting=\frac{84\times104.9}{111.7}=78.89g\:of\:CO\\$

Hence, 78.89 g of CO on reacting with excess of ferric oxide will produce 104.9 g of Fe.

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3 years ago

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Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

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3 years ago

How many grams of KCN are in 10.0 ml of a 0.10 M solution?

attashe74 [19]

Explanation:

As it is known that molarity is the number of moles present in a liter of solution.

Mathematically, Molarity = $\frac{no. of moles}{Volume in liter}$

As it is given that molarity is 0.10 M and volume is 10.0 ml. As 1 ml equals 0.001 L. Therefore, 10.0 ml will also be equal to 0.01 L.

Hence, putting these values into the above formula as follows.

Molarity = $\frac{no. of moles}{Volume in liter}$

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no. of moles = 0.001 mol

As molar mass of KCN is equal to 65.12 g/mol. Therefore, calculate the mass of KCN as follows.

No. of moles = $\frac{mass}{molar mass}$

0.001 mol = $\frac{mass}{65.12 g/mol}$

mass = 0.06152 g

Thus, we can conclude that 0.06152 grams of KCN are in 10.0 ml of a 0.10 M solution.

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