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2 years ago

Match the description to the type of mixture.

Chemistry

1 answer:

Softa [21]2 years ago

7 0

Answer:

1-homo

2-hetero

3-hetero

4-homo

Explanation:

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A solution of sulfuric acid has a concentration of 0.0980 g/L. If the density of the acid is 1.84 g/mL, what is the concentratio

mariarad [96]

The answer is 98ppm.

The ppm (Parts per million) is also a concentration unit. 1 ppm is equivalent to 1mg/L
Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
We rewrite, 0.0980 g = 0.0980*1000 mg = 98mg
Therefore, the concentration of the sulfuric acid solution is 98 mg/L = 98 ppm.

6 0

3 years ago

Read 2 more answers

What is the valency of aluminium and nitrate in Al (NO3)3

VikaD [51]

Answer:

Aluminum loses three valence shell electrons to become the aluminum ion Al+3. Nitrate is a polyatomic ion with the formula NO3 -1.

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2 years ago

According to the Energy Information Administration (EIA), oil accounts for what percentage of the U.S. energy mix?

stich3 [128]

Answer: answer is b 29%

Explanation:

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2 years ago

How many molecular of H2O and O2 are present in 8.5g of H2O2 ?

Mama L [17]

2H2O+O2--->2H2O2

8.5 gm H2O2=0.25 mole

hence H2O is also 0.25 mole i.e.4.5 gm

O2is 0.125 mole i.e.4 gm

4 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

EleoNora [17]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

4 0

2 years ago