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3 years ago

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What is the basic difference between exergonic and endergonic reactions? Group of answer choices Exergonic reactions release ene

rgy; endergonic reactions absorb it. Exergonic reactions involve ionic bonds; endergonic reactions involve covalent bonds. Exergonic reactions involve the breaking of bonds; endergonic reactions involve the formation of bonds. In exergonic reactions, the reactants have less chemical energy than the products; in endergonic reactions, the opposite is true.

1 answer:

alekssr [168]3 years ago

3 0

Answer: the basic difference is Exergonic reactions release energy and an endergonic reactions absorb energy .

HOPE THIS HELPS!!!

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CaCO3(s) CaO(s) + CO₂(g)

tatuchka [14]

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1 year ago

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

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2 years ago

Lab: Magnetic and Electric Fields Step 3: Determine the Polarity of Magnets

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Answer:

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Explanation:

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3 years ago

What is the molarity of a solution made by mixing 50.0 g of magnesium nitrate, Mg(NO3)2, in enough water to make 250. mL of solu

Ira Lisetskai [31]

............................

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3 years ago

A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25°C. What is the mole fraction of NaCl in this solu

Debora [2.8K]

Answer:

Mole fraction of Nacl is 0.173

Explanation:

we know that

$P_{sol}=\chi_{solvent}P^0_{solvent}$

where,

P sol - the vapor pressure of the solution

χ solvent - the mole fraction of the solvent

P ∘ solvent - the vapor pressure of the pure solvent

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at

25 ° C You can use an online calculator to find that the vapor pressure of pure water at 25 C is equal to about 23.8 torr .

$\chi_{water}= \frac{P{sol}}{P^0{water}}$

$\chi_{water}= \frac{19.6}{23.8}$

=0.827

Also we know that

$\chi_{water}+\chi_{Nacl}= 1$

This means that the mole fraction of sodium chloride is

χ_{Nacl}= 1-Χ_{water}

= 1-0.827 =0.173

3 0

3 years ago