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3 years ago

12

A solution is made by dissolving 15.0 mL of oil in enough gasoline to give 50.0 mL of solution. What is the % (v/v) of oil in th

e solution

1 answer:

Nastasia [14]3 years ago

4 0

Answer:

30% v/V

EXPLANATION

15 ml of oil in 50 ml of solution is 15/50 =30/100 =30% v/V

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago

A sample of water at 20 degrees c contains what bonds

vladimir2022 [97]

Covalent and hydrogen bonds

3 0

3 years ago

Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles o

Harman [31]

First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹

The answer is C

8 0

4 years ago

PLZ HELP Question 14 of 25 What is the name for a representation of the physical world?

JulijaS [17]

Answer:

Model

Explanation:

A model of anything is something you make to represent it in it's physical world form

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3 years ago

If 20.0 grams of Al is placed into a solution containg

meriva

Answer:

m H2(g) = 2.241 g H2(g)

Explanation:

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

limit reagent:

∴ Mw Al = 26.982 g/mol

∴ Mw H2SO4 = 98.0785 g/mol

⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al

⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4

⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2

∴ Mw H2 = 2.016 g/mol

⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2

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3 years ago