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3 years ago

Given the following equation: K+Cl2>KCl balance the equation

Chemistry

1 answer:

Sonja [21]3 years ago

8 0

Balanced equation : 2K + Cl2 > 2KCL.

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Gre4nikov [31]

Answer:

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Explanation:

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3 years ago

12 the products of the neutralization reaction between hydrochloric acid and magnesium hydroxide are

olga2289 [7]

The products of the neutralization reaction between hydrochloric acid and magnesium hydroxide are Magnesium chloride and water.

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3 years ago

How many grams of C2H6 are needed to react with 54 grams of O2?

Dmitrij [34]

Answer:

Explanation:

in a combustion of ethane 2 moles of ethane react with 7 moles of O2

now no of moles in 54 gram of O2=mass/ molar mass

moles =54/32=1.7 moles

if 7 moles of O2 required 2 moles of ethane then 1.7 mole required=?

7 moles of O2=2 moles of C2H6

1.7 moles of O2=1.7*2/7=0.5 moles of C2H6

0.5 moles of C2H6 contain how much grams=?

mass= moles*molar mass=0.5*30=15

4 0

3 years ago

What is the percent composition by mass of aluminum in Al2(SO4)3 (gram-formula mass = 342 grams/mole)?

Aliun [14]

The atomic mass of Aluminum is around 27 amu. There are two aluminum atoms, so it takes 54 amu. The gram-formula states that 1 mole of this compound has 342 grams. By definition, this means that the total amu of this compound is 342. Therefore, we divide 54 by 342 to get around 15.8 %.

8 0

3 years ago

Read 2 more answers

g 304 mL of a 0.36 M potassium hydroxide solution is added to 341 mL of a 0.51 M lithium hydroxide solution. Calculate the pOH o

Liono4ka [1.6K]

Answer:

pOH = 0.36

Explanation:

Both potassium hydroxide and lithium hydroxide solutions are strong bases, so you assume 100% dissociation.

1. Potassium hydroxide solution, KOH

Volume, V = 304 mL = 0.304 liter

Molarity, M = 0.36 M

number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol

1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944

2. LIthium hydroxide, LiOH

Volume, V = 341 mL = 0.341 liter

Molarity, M = 0.51 M

number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol

1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391

3. Resulting solution

Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol

Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter

Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M

4. pOH

$pOH=-\log [OH^-]=-\log (0.439)=0.36$ ← answer

5 0

3 years ago