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Sauron [17]
3 years ago
10

What volume of a 1.0 M HCl is required to completely neutralize 25.0 ml of a 1.0 M KOH?

Chemistry
1 answer:
kirza4 [7]3 years ago
8 0
The question above can be solved by using this equation: 
CAVA =CBVB
Where:
CA =Concentration of acid = 1.0 M
VA = Volume of acid = ?
CB = Concentration of base = 1.0 M
VB = Volume of base = 25 ml
VA = CBVB / CA
VA = [1 * 25] / 1 = 25 / 1 = 25
VA = 25 ml
Therefore, the volume of acid that is required to completely neutralize the base is 25 ml.<span />
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One of the examples of Boyles law in action is a syringe 




3 0
3 years ago
What is the mass of 7.50 moles of magnesium chloride, mgcl2?
Natali5045456 [20]
MgCl2 = 1Mg + 2Cl = 1(24.3) + 2(35.45) = 95.2g/1mole
7.50moles MgCl2 x 95.2g MgCl2 = 714g MgCl2
8 0
3 years ago
Which statements describe band theory? Check all that apply.
Papessa [141]

Answer:

b c

Explanation:

right

8 0
3 years ago
9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
2 years ago
The density of pure solid copper is 8.94 g/mL. What volume does 34 g of copper occupy?
patriot [66]

hey there ! :

Density = 8.94 g/mL

mass = 34 g

Volume = ??

Therefore:

D = m / V

8.94 = 34 / V

V = 34 / 8.94

V = 3.803 mL

Hope this helps!

4 0
3 years ago
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