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3 years ago

The number of grams in the molar mass of an element is the same as an elements ...

Chemistry

1 answer:

rosijanka [135]3 years ago

4 0

It is the same as an elements number of protons.

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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 $C_3H_8(g)$

7 0

2 years ago

If this container of liquid water was frozen into ice and then, thawed back into liquid what would the volume be?

Dmitry_Shevchenko [17]

Answer:

Explanation:

because it's is still the same amount even if you freeze it

6 0

2 years ago

Can someone help me with #35,37,39,41 please

adoni [48]

35 b.
37 h
39 i
41 d
hope that helps

4 0

2 years ago

Find the percent composition of water in barium chloride dihydrate. Round your answer to the

nadya68 [22]

Answer: its 15 its none of those

Explanation:

7 0

1 year ago

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the

Fofino [41]

Answer:

(a) $102.6g/mol$

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

$V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}$

Solving for the molarity of base we've got:

$M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L$

Now, we can compute the moles of the base as:

$n_{base}=0.425mol/L*0.1L=0.0425mol$

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

$\frac{4.36g}{0.0425mol} =102.6g/mol$

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

$102.6g/mol-16g/mol-1g/mol=85.6g/mol$

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0

2 years ago