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2 years ago

A compound contains 40%carbon, 6.7%hydrogen, and 56.3%oxygen

Chemistry

1 answer:

ahrayia [7]2 years ago

7 0

180 might be the answer man

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Calculate the missing variables in each experiment below using Avogadro’s law.

blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml n1 = 9.11 mol

V2 = 123 ml n2 = ?

Formula

$\frac{V1}{n1} = \frac{V2}{n2}$

$n2 = \frac{n1V2}{V1}$

$n2 = \frac{(9.11)((123)}{(612)}$

n2 = 1.83 mol

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3 years ago

The percent composition of calcium is ?

DochEvi [55]

Answer: 40.1%

Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.

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4 years ago

At a particular temperature, K 3.39 for the reaction SO2(9) + NO2(9) SOs(9) +NO(9) If all four gases had initial concentrations

Korvikt [17]

Answer:

The equilibrium concentrations are:

[SO2]=[NO2] = 0.563 M

[SO3]=[NO] = 1.04 M

Explanation:

<u>Given:</u>

Equilibrium constant K = 3.39

[SO2] = [NO2] = [SO3] = [NO] = 0.800 M

<u>To determine:</u>

The equilibrium concentrations of the above gases

Calculation:

Set-up an ICE table for the given reaction

$SO2(g) + NO2(g)\rightleftharpoons SO3(g) + NO(g)$

I 0.800 0.800 0.800 0.800

C -x -x +x +x

E (0.800-x) (0.800-x) (0.800+x) (0.800+x)

The equilibrium constant is given as:

$Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}$

$3.39=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}$

x = 0.2368 M

[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M

[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M

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3 years ago

What is the definition of DNA in meiosis

Eddi Din [679]

In meiosis, DNA replication is followed by two rounds of cell division to produce four daughter cells, each with half the number of chromosomes as the original parent cell. The two meiotic divisions are known as meiosis I and meiosis II.

Explanation:

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3 years ago

How many moles of O2 molecules should be supplied to burn 1 mol of CH4 molecules in a domestic furnace?

Sati [7]

The combustion of 1 mole of methane (CH4) in a domestic furnace requires 2 moles of O2 molecules, assuming the combustion was complete or ideal. To solve this problem, use stoichiometry of the reaction's balanced chemical equation:

CH4 + 2O2 --> CO2 + 2H2O

The ratio of CH4 to O2 in terms of moles is 1:2. So 1 mole of CH4 needs 2 moles of O2.

4 0

4 years ago