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3 years ago

Determine the number of grams of Carbon in 215 g of sucrose.

Chemistry

1 answer:

olya-2409 [2.1K]3 years ago

6 0

just find the percent mass of oxygen in sucrose again. and then multiply that by 50.00.

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3 years ago

Answer truthfully:))

Elina [12.6K]

Answer:

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3 years ago

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What is oxidation number of S in H2SO5??

mote1985 [20]

★ « what is oxidation number of S in H2SO5?? » ★

it's 6!!

Explanation:

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2 years ago

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

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3 years ago

Select the single best answer. Which mechanism properly shows the movement of electrons in the reaction? 2xsafari

sweet [91]

Answer:

Explanation:

We must study the reaction pictured in the question closely before we begin to attempt to answer the question.

Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.

Hence we must show electron movements in both species using a half arrow.

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3 years ago