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777dan777 [17]
4 years ago
11

A 0.229-g sample of an unknown monoprotic acid is titrated with 0.112 m naoh. the resulting titration curve is shown here. part

a determine the molar mass of the acid. express your answer using two significant figures.
Chemistry
1 answer:
Kazeer [188]4 years ago
4 0
It would be 0.341 because if you add 0.229 and 0.112 it will be 0.341
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Will mark brainliest!!
olga2289 [7]

Answer:

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6 0
3 years ago
Natural gas is a mixture of many substances, primarily CH₄, C₂H₆, C3H8, and C4₄H₁₀. assuming that the total pressure of the gase
olchik [2.2K]

Answer:

A. The partial pressure for CH4 = 0.0925atm

B. The partial pressure for C2H6 = 0.925atm

C. The partial pressure for C3H8 = 0.346atm

D. The partial pressure for C4H10 = 0.115atm

Explanation:

Total pressure = 1.48atm

Total mole = 0.4+4+1.5+0.5=6.4

A. Mole fraction of CH4 = 0.4/6.4 = 0.0625

The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm

B. Mole fraction of C2H6 = 4/6.4 = 0.625

The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm

C. Mole fraction of C3H8 = 1.5/6.4 = 0.234

The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm

D. Mole fraction of C4H10 = 0.5/6.4 = 0.078

The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm

7 0
3 years ago
What happens when the pressure of a gas is decreased?
Serggg [28]

Answer:

The combined gas law states that the pressure of a gas is inversely related to the volume and directly related to the temperature. If temperature is held constant, the equation is reduced to Boyle's law. Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase.

Explanation:

3 0
3 years ago
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drek231 [11]
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6 0
3 years ago
Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
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