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2 years ago

Assign the oxidation number for (NH4)3PO4

2 answers:

6 0

N(-3)
N(+1)
P(+5)
O(-2)

-3*3+(+1)*12+(+5)+(-2)*4=-9+12+5-8 =0 -the sum of oxidation number in the molecule should be equal zero.

baherus [9]2 years ago

3 0

Explanation:

Hello!

Let's solve this!

In a salt, we must take into account that the sum of the oxidation numbers is equal to zero since the molecule is stable.

In this case the molecule is a salt and is (NH4) 3PO4

N = -3

H = + 1 (there are four atoms in total is +4)

(NH4) 3 = (- 3 + 4) * 3 = + 3

P = + 5

O = -2 (since there are 4 atoms in total it is -8)

PO4 = + 5-8 = -3

The total is: -3 + 3 = 0

We verify that the charges give a total of zero, so that our calculation is well done.

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6 months ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<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>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

2 years ago