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Scorpion4ik [409]
3 years ago
9

Containers of ice cream are sold by volume, not by weight. Some ice cream manufacturers pump air into the ice cream before packa

ging it. Why does this practice make the ice cream more expensive to the consumer?
A) Air weighs more than ice cream does.
B) Air weighs less than ice cream does.
C) After air is pumped into ice cream, the mixture occupies a greater volume than the ice cream alone did.
D) After air is pumped into ice cream, the mixture occupies a smaller volume than the ice cream alone did.
Chemistry
1 answer:
Arturiano [62]3 years ago
7 0
C. Adding air would add to volume. Since it's by volume not weight they'd want more volume because more volume means more money
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What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..
Aloiza [94]

Molar mass of NH_3

\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

No of moles=Given mass/Molar mass

\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

Now

Lets write the balanced equation

\\ \sf\longmapsto N_2+3H_2=2NH_3

  • There is 2moles of Ammonia
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  • 1mole of N_2

Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

For Hydrogen

\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

\\ \sf\longmapsto 1.8\times 10^{22}molecules

For Ammonia

\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

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For Nitrogen

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3 0
2 years ago
How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?
Nata [24]

Answer:

\boxed{\text{53 g }}

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r:                                 142.04  

             2NaOH + … ⟶ Na₂SO₄ + …  

n/mol:      0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces \boxed{\text{53 g }} of Na₂SO₄.

4 0
3 years ago
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