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Mekhanik [1.2K]
2 years ago
6

I set up the conversion factor as 1 mol H20 / 18.02 g H2O because?

Chemistry
1 answer:
Andrej [43]2 years ago
8 0

Answer:

How many grams of H2O are in 1.0 mole of H2O?

18.02 grams

The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams.

#Yourchuu

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Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to ________.
kkurt [141]

Answer : Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to volume.

Explanation :

Weight by volume (w/v) means that the mass of solute present in 100 mL volume of solution.

Weight by weight (w/w) means that the mass of solute present in 100 gram of solution.

Volume by volume (v/v) means that the volume of solute present in 100 mL volume of solution.

As per question, amoxicillin suspension is, 125 mg/ 5 ml that means 125 mg of Amoxicillin present in 5 mL of suspension. So, it is an example of weight to volume.

Hence, it is an example of weight to volume.

8 0
3 years ago
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olganol [36]

Answer:

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Explanation:

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8 0
2 years ago
Convert 1.25 atm to bar.
Tasya [4]

Answer: 1.27 bar

Explanation:

1 atm = 1.01325 bar

1.25 atm = Z (let Z be the unknown value)

To get the value of Z, cross multiply

Z x 1 atm = 1.25 atm x 1.01325 bar

1 atm•Z = 1.2665625 atm•bar

To get the value of Z, divide both sides by 1 atm

1 atm•Z/1 atm = 1.2665625 atm•bar/1atm

Z = 1.2665625 bar

(Round up Z to the nearest hundredth as 1.27 bar)

Thus, 1.25 atm when coverted gives 1.27 bar

8 0
3 years ago
Find the ratio of the effusion rate of hydrogen gas to the effusion rate of krypton gas.
Zielflug [23.3K]

This problem could be solved through the Graham’s law of effusion (also known as law of diffusion). This law states that the ratio of the effusion rate of the first gas and effusion rate of the second gas is equivalent to the square root of the ratio of its molar mass. Thus the answer would be 0.1098. 

6 0
3 years ago
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A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature rose b
dezoksy [38]

Answer:

872.28 kJ/mol

Explanation:

The heat released is:

ΔH = C*ΔT

where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:

ΔH =  9.47*5.72 = 54.1684kJ

A quantity of 1.922 g of methanol in moles are:

moles = mass / molar mass

moles = 1.992/32.04 = 0.0621 mol

Then the molar heat of combustion of methanol is:

ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol

5 0
3 years ago
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