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3 years ago

Classify equation Write the correct chemical formula Balance equation

Chemistry

1 answer:

Tanya [424]3 years ago

6 0

You classify the equation as single replacement or double replacement. Single replacement has the form: A + BC --> AC + B. Double replacement has the form: AB + CD --> AD + BC.

1.
a. Single replacement
b. Al + Fe₂O₃ --> Al₂O₃ + Fe
c. 2Al + Fe₂O₃ --> Al₂O₃ + 2Fe

2.
a. Double replacement
b. Pb(C₂H₃O₂)₂ + H₂S --> PbS + HC₂H₃O₂
c. Pb(C₂H₃O₂)₂ + H₂S --> PbS + 2 HC₂H₃O₂

3.
a. Single Replacement
b. Mg + HCl --> MgCl₂ + H₂
c. Mg + 2HCl --> MgCl₂ + H₂

4.
a. Single Replacement
b. Zn + Pb(NO₃)₂ --> Zn(NO₃)₂ + Pb
c. Zn + Pb(NO₃)₂ --> Zn(NO₃)₂ + Pb

5.
a. Double replacement
b. AgNO₃ + NaCl --> AgCl + NaNO₃
c. AgNO₃ + NaCl --> AgCl + NaNO₃

6.
a. Single replacement
b. K + Ba(C₂H₃O₂)₂ --> KC₂H₃O₂ + Ba
c. 2 K + Ba(C₂H₃O₂)₂ --> 2 KC₂H₃O₂ + Ba

7.
a. Double replacement
b. LiOH + Fe(NO₃)₃ --> Fe(OH)₃ + LiNO₃
c. 3 LiOH + Fe(NO₃)₃ --> Fe(OH)₃ + 3 LiNO₃

8.
a. Single replacement
b. CuO + H₂ --> Cu + H₂O
c. CuO + H₂ --> Cu + H₂O

9.
a. Single Replacement
b. K + H₂O --> KOH + H₂
c. 2 K + 2 H₂O --> 2 KOH + H₂

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Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05 $\frac{g}{mol}$ .

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · $\frac{1 mol Na^2SO^4}{142.05g Na^2SO^4}$ · $\frac{4 mol O}{1 mol Na^2SO^4}$ · $\frac{6.022x10^2^3}{1 mol O}$

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What is the molarity of the Ca(OH)2 solution if 32.00 mL of Ca(OH)2 requires 16.08 mL of a 2.303 M solution for complete titrati

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Answer:

For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb

Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L

So now we can plug into the equation and solve:

1.338 mol/L * 0.01875 L= x mol/L * 0.02473 L

This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.

First step: setup the stoichiometry problem:

3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl

Second step: Molarity equation

Molarity= moles HCl/ L M= mol HCl/0.04027 L

For the third problem, you will just use the same equation as the first: MaVa= MbVb

Ma=0.57 M Va= x L Mb= 0.875 M Vb= 23.83 mL= 0.02383 L

0.57 M * x L= 0.875 M * 0.02383 L

With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.

0.75 M= x mol/ 0.0227 L mol NaOH = 0.75 M *0.0227 L mol NaOH= 0.017025

So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles

Molar Mass= 3.6 g/ 0.017025 mol

And with that you will find the molar mass of HX, and even determine what X is.

Explanation:

hope I helped

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