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3 years ago

Need done ASAP HELP MEEE

Chemistry

1 answer:

blsea [12.9K]3 years ago

5 0

Answer:

1. Heat

2. Temperature

3. convection

4. Conduction

5. Radiation

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Helium gas diffuses 4 times as fast as an unknown gas. What is relative molecular mass of the gas

Inessa05 [86]

The relative molecular mass of the gas : 64 g/mol

<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

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$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

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2 years ago

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C

Kipish [7]

Answer: 72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

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

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

Putting in the values we get:

$\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles$

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

According to stoichiometry:

1 mole of $C_3H_8$ requires 5 moles of oxygen

0.45 moles of $C_3H_8$ require= $\frac{5}{1}\times 0.45=2.25$ moles of oxygen

Mass of $O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g$

72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

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3 years ago