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Elis [28]
3 years ago
13

Name the 6 kinetic factors

Chemistry
1 answer:
ryzh [129]3 years ago
5 0
Kinetics is the study of reaction rates and how they are affected. Many factors, such as concentration, pressure, temperature, and enzyme activity, can impact the rate of a reaction.
You might be interested in
A sample of methane (CH4) has a volume of 25 mL at a pressure of 0.80 atm. What is the volume of the gas at each of the followin
Xelga [282]

Answer:

a. 50ml b.10ml c. 6.097ml d. 190.1 ml

Explanation:

According to Boyle's law

Volume is inversely proportional to pressure at constant temerature

Mathematically

P1V1=P2V2

P1=Initial pressure=0.8atm

V1=Initial volume=25ml

making V2 the subject

at 0.4atm P2=0.4 atm,

V2=25×0.8/0.4

=50ml

at 2 atm V2=25×0.8/2

=10 ml

1mmHg=0.00131579

2500mmHg=3.28 atm

At 3.28 atm,V2=25×0.8/3.28

=6.097 ml

at 80.0 torr

1 torr=0.00131579

80 torr=0.1052 atm

at 0.1048 atm V2=25×0.8/0.1048

=190.1 ml

4 0
3 years ago
What is the mass of 2.50 moles of NaCl
garik1379 [7]

Answer:

The mass of 2,50 moles of NaCl is 146, 25 g.

Explanation:

First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:

Weight NaCl= Weight Na + Weight Cl=  23 g+ 35,5 g= 58, 5 g/ mol

1 mol ------ 58, 5 g

2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>

8 0
3 years ago
How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

Learn more about precipitation here:

brainly.com/question/13859041

#SPJ4

6 0
2 years ago
a propane torch is lit inside a hot air balloon during preflight preparations to inflate the balloon. which condition of the gas
earnstyle [38]

Answer:

The pressure remains constant

Explanation:

this is an example in charles law where as the temperature increases so does the volume.

5 0
3 years ago
Read 2 more answers
A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958
Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

<u>Step 1</u>: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

<u>Step 2</u>: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

<u>Step 3:</u> Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

<u>Step 4:</u> Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

<u>Step 5:</u> Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

<u>Step 6:</u> Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

<u>Step 7: </u>Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol  

<u>Step 8:</u> Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

<u>Step 9</u>: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

<u>Step 10</u>: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0
3 years ago
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