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3 years ago

12

PLZ HELP WILL BRAINLIEST

1 answer:

finlep [7]3 years ago

7 0

Answer:

6,3,2

Explanation:

Fe2(SO4)3 + 6 KOH = 3 K2SO4 + 2 Fe(OH)3

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Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

In 1865 a pond was surrounded by open fields today the same area is swampy and surrounded by a forest which process is responsib

Nookie1986 [14]

Answer:

erosion

Explanation:

3 0

3 years ago

Read 2 more answers

Is bleach liquid starch? <br> Yes or No

Jobisdone [24]

Answer:

O it's not

Explanation:

Have a great day!

8 0

3 years ago

Zinc metal reacts with copper sulfate through the following

morpeh [17]

Answer:

26.9

Explanation:

5 0

3 years ago

The empirical formula of a compound is CH2O and its molecular weight is 90.09 g/mol. Determine the molecular formula.

Pachacha [2.7K]

Answer : The molecular formula of the compound will be, $C_{3}H_{6}O_3$

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

As we are given that the empirical formula of a compound is $CH_2O$ and the molar mass of compound is, 90.09 gram/mol.

The empirical mass of $CH_2O$ = 1(12) + 2(1) + 1(16) = 30 g/eq

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

$n=\frac{90.09}{30}=3$

Molecular formula = $(CH_2O)_n=(CH_2O)_3=C_{3}H_{6}O_3$

Thus, the molecular formula of the compound will be, $C_{3}H_{6}O_3$

7 0

4 years ago