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2 years ago

7. A. How many moles of Calcium hydroxide (CaOH2) are needed to completely react with 14.5 mol of Phosphoric acid (H3PO4)?

Chemistry

1 answer:

Artyom0805 [142]2 years ago

6 0

Answer:

moles of calcium hydroxide= 21.75 mol

a) 43.5 mol

b) 7.25 mol

Explanation:

Please see the attached picture for the full solution.

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Which of these elements has an atom with the most stable outer electron configuration? * Cl Ca Ne Na

SashulF [63]

Answer:

Neon (Ne) has the most stable outer electron configuration because the outer electron is completely filled and it has octet structure

Explanation:

The configuration of these elements is as follows;

Cl₁₇ = 2, 8,7 (the outer electron is 7)

Ca₂₀ = 2,8,8,2 (the outer electron is 2)

Ne₁₀ = 2,8 (the outer electron is 8)

Na₁₁ = 2,8,1 (the outer electron is 1)

Based on the outer electron value above, Neon (Ne) has the most stable outer electron configuration because the outer electron is completely filled and it has octet structure.

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3 years ago

A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.

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3 years ago

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Which of the reactants or products is a gas<br><br> c4h10<br><br> o2<br><br> co2

Alex

Answer:

co2

Explanation:

because carbon is a gas product

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2 years ago

How many moles are in 3.24 x 1020 particles of lead? (show work plz)

lesya [120]

<h3>Answer:</h3>

0.000538 mol Pb

<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>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.24 × 10²⁰ particles Pb (lead)

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.24 \cdot 10^{20} \ particles \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ particles \ Pb})$
Multiply/Divide: $\displaystyle 0.000538 \ mol \ Pb$