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Sidana [21]
3 years ago
13

All of the following are exothermic reactions except,

Chemistry
1 answer:
monitta3 years ago
8 0
Exothermic reactions release heat the surrounding. So Freezing? 
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A gas of unknown molar mass was allowed to effuse through a small opening under constant pressure conditions. It required 986 s
Kobotan [32]

Answer:

The molar mass of the unknown gas is 154.4 g/mol

Explanation:

Step 1: Data given

It takes 986 seconds for an unknown gas to effuse

It takes 811 seconds for chlorotrifluoromethane (CClF3) gas to effuse

Molar mass of chlorotrifluoromethane (CClF3) = 104.46 g/mol

Step 2: Calculate the molar mass of the gas

t1/t2/ √(M1/M2)

⇒with t1 = the time needed for the unknown gas to effuse = 986 seconds

⇒with t2 = the time needed for CClF3 to effuse = 811 seconds

⇒with M1 = the molar mass of the unknown gas = TO BE DETERMINED

⇒with M2 = the molar mass of CClF3 = 104.46 g/mol

986/811 = √(M1/104.46)

(986/811)² = M1 / 104.46

M1 = 154.4 g/mol

The molar mass of the unknown gas is 154.4 g/mol

6 0
3 years ago
Jdbdjevejeveheveveebdbdhdhdhd
ahrayia [7]

Answer:

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Explanation:

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5 0
2 years ago
Read 2 more answers
Difference between hard water and cold water ? ​
sweet-ann [11.9K]
The hardness of water is determined primarily by the amount of calcium and magnesium it contains. ... Instead of having higher levels of calcium and magnesium, soft water tends to have higher concentrations of sodium, or salt.
6 0
3 years ago
Color change is most indicative of which type of chemical reaction?
makvit [3.9K]

Answer; chemical reaction

4 0
3 years ago
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-1
Irina-Kira [14]

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

1 = \%_{16} + \%_{17} + \%_{18}

1 = x + 3.2 \cdot 10^{-4} + \%_{18}

\%_{18} = 1 - x - 3.2 \cdot 10^{-4}

Hence, the percent abundance of O-18 is:  

m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}  

15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)

x = 0.980614 \times 100 = 98.0614 \%                                                              

Hence, the percent abundance of oxygen-18 is:

\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

8 0
3 years ago
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