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scoray [572]
3 years ago
15

I need help, do you mind helping me?

Chemistry
1 answer:
Sever21 [200]3 years ago
5 0
The mole ratio is 2:19 or 1:9.5
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Meiosis is a process where a single cell divides twice to produce four cells containing half the original amount of genetic information. These cells are our sex cells – sperm in males, eggs in females. ... These four daughter cells only have half the number of chromosomes? of the parent cell – they are haploid.

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Molten zinc chloride can be electrolysed.
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Answer: The products of electrolysis are Zn(l) and Cl_2 gas

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Electrolysis of a subastance is breaking it into its constituents by the action of electrical current.  

In the electrolysis of molten zinc chloride, zinc metal is produced at the cathode which is a negative electrode and chlorine gas produces as the anode which is a positive electrode.

ZnCl_2(l)\rightarrow Zn^{2+}+2Cl^-

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3 years ago
Fluorine gas reacts violently with water to produce hydrogen fluoride and ozone according to the following unbalanced equation.
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Answer:

Fluorine gas reacts violently with water to produce hydrogen fluoride and ozone according to the following unbalanced equation. How many moles of o, are created assuming STP conditions are in effect, given

Explanation:

3

6 0
3 years ago
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
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