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3 years ago

Help I need to show work!

Chemistry

1 answer:

lubasha [3.4K]3 years ago

6 0

Answer:

a. Combustion.

b. $m_C=3.057gC$

c. $m_H=3.67gH$

Explanation:

Hello.

a. In this case, given the reaction by which propane is converted into carbon dioxide and water:

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

It is known as a combustion reaction since it is about a fuel (here propane) which is burnt by the oxygen contained in the air (21%).

b. Since the carbon dioxide contains all the carbon in the products based on the law of conservation of mass, the yielded grams are computed via a mole mass relationship:

$m_C=11.21gCO_2*\frac{12gC}{44gCO_2}=3.057gC$

Since 44 grams of carbon dioxide contain 12 grams of carbon.

c. As well as b., all the hydrogen is given off in form of water, thus, the required mass turns out:

$m_H=33.0gH_2O*\frac{2gH}{18gH_2O}=3.67gH$

Since 18 grams of water contain 2 grams of hydrogen.

Best regards.

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A chemist requires 5.00 liters of 0.420 M H2SO4 solution. How many grams of H2SO4 should the chemist dissolve in water? 129 gram

natali 33 [55]

Answer:

Approximately 206 grams.

Explanation:

How many moles of sulfuric acid $\mathrm{H_2SO_4}$ are there in this solution?

$\text{Number of moles of solute} = \text{Concentration} \times \text{Volume}$ .

The unit for concentration " $\mathrm{M}$ " is equivalent to mole per liter. In other words, $\rm 1\;M = 1\; mol\cdot L^{-1}$ . For this solution, the concentration of $\mathrm{H_2SO_4}$ is $\rm 0.420\;M = 0.420\; mol\cdot L^{-1}$ .

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V\\&= \rm 0.420\;mol\cdot L^{-1}\times 5.00\; L \\&= \rm 2.10\; mol\end{aligned}$ .

What's the mass of that $\rm 2.10\; mol$ of $\mathrm{H_2SO_4}$ ?

Start by finding the molar mass $M$ of $\mathrm{H_2SO_4}$ .

Relative atomic mass data from a modern periodic table:

H: 1.008;
S: 32.06;
O: 15.999.

$\displaystyle M(\mathrm{H_2SO_4}) = 2\times \underbrace{1.008}_{\mathrm{H}} + 1\times \underbrace{32.06}_{\mathrm{S}} + 4\times \underbrace{15.999}_{\mathrm{O}} = \rm 98.072\;g\cdot mol^{-1}$ .

$\text{Mass} = \text{Quantity in moles} \times \text{Molar Mass}$ .

$m = n \cdot M = \rm 2.10\; mol \times 98.072\;g\cdot mol^{-1} \approx 206\; g$ .

In other words, the chemist shall need approximately 206 grams of $\mathrm{H_2SO_4}$ to make this solution. As a side note, keep in mind that the 206 grams of $\mathrm{H_2SO_4}$ also take up considerable amount of space. Therefore it will take less than 5.00 L of water to make the 5.00 L solution.

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4 years ago

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3 years ago

Perform each of the following unit conversions using the conversion factors given below: 1 atm = 760 mmHg = 101.325 kPa

Alenkasestr [34]

unit coversation
1.429 atm - 1086mmhg

9361 pa-9.36 KPa - 70.21 mmhg

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calculation

(a) 1 atm = 760 mmhg
1.429 atm = ?
1.429 x760/1 = 1086.34 mm hg

(B) 1 mmhg = 101.325 kpa
? =9361 KPa
9361 x1 /101.25 =70.21 mmhg

760 mm hg= 101.325 KPa
70.21 mm hg=?

70.21 x101.325/760 = 9.36 Kpa

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? = 725
= 725 x1/ 760=0.95 atm

1 atm = 101.325 kpa
0.95 =?

0.95 x101.325/1 = 96.26 KPa

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