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3 years ago

20 points!!!

1 answer:

3 0

A)
Each mole of Rb₃PO₄ will release:
3 moles of Rb⁺
1 mole of PO₄⁻³
So a total of 4 moles of ions are released per mole.
If there are 0.5 moles, 2 moles of ions will be released.

B)
Each mole of Cr₂(SO₄)₃ will release:
2 moles of Cr⁺³
3 moles of SO₄⁻²

So a total of 5 moles of ions are released per mole.
When there are 0.5 moles, 2.5 moles of ions will be released.

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What would be the correct Keq expression for the following reaction that has reached dynamic equilibrium?

Katyanochek1 [597]

So we have an equilibrium expression:
CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

The equilibrium expression is the concentration of each product (excluding solids and liquids in all cases) raised to the power of their coefficient, all divided by the concentration of each reactant (again, excluding solids and liquids). Noting that brackets around a molecule indicate concentration, the expression would be:
([CH4]*[H2O])/([CO]*[H2]^3)

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3 years ago

Name all of your mha ships and i will draw them!

malfutka [58]

Answer:

I don't really ship anyone but if I had to pick it would probably be either kamijiro or tododeku.

Explanation:

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3 years ago

What is the mass of 1.58 moles of CH4

HACTEHA [7]

<h3>Answer:</h3>

25.4 g CH₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.58 mol CH₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})$
Multiply/Divide: $\displaystyle 25.359 \ g \ CH_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

25.359 g CH₄ ≈ 25.4 g CH₄

4 0

3 years ago

How many atoms of oxygen are in 1.0 g of k2o2 (mw 110.0)?

sasho [114]

Given

Mass of K2O2 = 1.0 g

Molar mass of K2O2 = 110.0 g/mol

Calculate the # moles of K2O2

# moles = mass of K2O2/molar mass

= 1.0 g/110.0 g.mol-1 = 0.0091 moles

1 mole of K2O2 contains 2*6.023*10^23 atoms of oxygen

Therefore, 0.0091 moles of K2O2 will correspond to:

= 0.0091 moles * 2* 6.023*10^23 atoms/1 mole

= 1.096 * 10^24 atoms of oxygen

5 0

3 years ago

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Dovator [93]

Answer:

a and d

Explanation:

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3 years ago