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denis-greek [22]

3 years ago

14

Properties of gases are dependent upon the pressure, temperature, and volume of the container they are in, as well as the number

of gas particles. true or false

1 answer:

OLga [1]3 years ago

5 0

Answer:

False

Explanation:

Those would affect physical properties not chemical properties.

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Which type of mixture can be separated using a sieve?

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FILTRATION When the substances in a mixture have different particle sizes, they are separated by filtration. The mixture is poured through a sieve or filter. The smaller particles slip through the holes, but the larger particles do not.p

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Is this true or false? :D

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Which metal is heaviest ?

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4 years ago

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A solution contains 0.0330 M Pb2 (aq) and 0.0330 M Sr2 (aq). If we add SO42–(aq), what will be the concentration of Pb2 (aq) whe

Fiesta28 [93]

The Ksp of PbSO₄ = 2.53 × 10⁻⁷
for srSO₄→ 3.44 × 10⁻⁷
for srSO₄
srSO₄→ Sr²⁺ + SO₄⁻
Ksp = [sr²⁺] [SO₄⁻]
3.44 ˣ 10⁻⁷ = (0.0330M) [ SO₄⁻]
[SO₄⁻] = 8.1 ˣ 10⁻⁶M
foe Pb SO₄
PbSO₄ →Pb²⁺ + SO₄⁻
Ksp = [ Pb²⁺] [ SO₄⁻]
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6 0

4 years ago

Consider the reaction of C3H8 with O2 to form CO2 and H2O. If 5.11 g O2 is reacted with excess C3H8 and 3.35 g of CO2 is ultimat

JulsSmile [24]

Answer : The percent yield of the reaction is, 79.8 %

Explanation : Given,

Mass of $O_2$ = 5.11 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $O_2$ .

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{5.11g}{32g/mole}=0.159mole$

Now we have to calculate the moles of $CO_2$ .

The balanced chemical reaction will be,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction, we conclude that

As, 5 moles of $O_2$ react to give 3 moles of $CO_2$

So, 0.159 moles of $O_2$ react to give $\frac{3}{5}\times 0.159=0.0954$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.0954mole)\times (44g/mole)=4.1976g$

The theoretical yield of $CO_2$ = 4.1976 g

The actual yield of $CO_2$ = 3.35 g

Now we have to calculate the percent yield of $CO_2$

$\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}\times 100=\frac{3.35g}{4.1976g}\times 100=79.8\%$

Therefore, the percent yield of the reaction is, 79.8 %

8 0

3 years ago