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stepan [7]
2 years ago
5

1. Explain: what is the purpose of balancing a chemical equation?

Chemistry
2 answers:
spayn [35]2 years ago
6 0
The purpose of balancing a chemical equation is to get the correct proportions of reagents and products for a reaction.

How to balance the Chemical Equation:
Step 1: count the number of atoms on each side of the equation
Step 2: Look at coefficients (the number In front of the element) and balance the equation
Step 3: ONLY CHANGE COEFFICIENTS
Step 4: Count polyatomic ions as one item
labwork [276]2 years ago
5 0
Answer for 1: The purpose of balancing a chemical equation is to obtain the correct products for a given reaction

Answer for 2:
Step 1: Write the unbalanced chemical equation

Step 2: Write down the number of atoms

Step 3: Add coefficients to balance mass in a chemical equation

Step 4: balance oxygen and hydrogen atoms
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How many total atoms are in 0.290 g of P2O5?
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There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\

= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\

Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

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8 0
3 years ago
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
muminat

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ? 

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

n = \dfrac{m}{MM}

5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}

m = 5.20*84.2

\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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