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2 years ago

1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the

Chemistry

1 answer:

borishaifa [10]2 years ago

8 0

A. 1
B. 4
C. 5
D. 3
E. 2

The correct answer is E. 2

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Find the mass of 175.4mL of benzene if the density is 0.8786g/ml

bulgar [2K]

Hey there!:

Volume = 175.4 mL

density = 0.8786 g/mL

mass = ?

Therefore:

D = m / V

0.8786 = m / 175.4

m = 0.8786 * 175.4

m = 154.10644 g

hope this helps!

5 0

3 years ago

Number of neutrons for 9BE4

GREYUIT [131]

Mass number - # of protons = #neutrons, so I would say the answer is 5

8 0

2 years ago

Scientific notation is used to express large numbers in convenient form true or false

Olin [163]

Scientific notation is used to express large numbers in a way that is to use, readable, comparable to other numbers and convenient. It is especially useful for things that are very large and very small because it is very tedious to work with many zero's when completing complex math problems.

4 0

3 years ago

A sample of nitrogen gas occupies a volume of 2.55 L when it is at 755 mm Hg and 23 degrees Celsius. Use this information to det

ivolga24 [154]

Answer: The number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

Explanation:

To calculate the amount of nitrogen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 2.55 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 2.55L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 296K\\\\n=\frac{755\times 2.55}{62.364\times 296}=0.1043mol$

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=755mmHg\\V_1=2.55mL\\T_1=23^oC=[23+273]K=296K\\P_2=?\\V_2=4.10L\\T_2=18^oC=[18+273]K=291K$

Putting values in above equation, we get:

$\frac{755mmHg\times 2.55L}{296K}=\frac{P_2\times 4.10L}{291K}\\\\P_2=\frac{755\times 2.55\times 291}{4.10\times 296}=461.6mmHg$

Hence, the number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

6 0

3 years ago

Read 2 more answers

In a fission reaction of U-235, there was a release of 9.20 × 1011 kJ of energy. What amount of mass in kilograms would have bee

Hatshy [7]

e = mc^2

m = e / c^2

Plug in your numbers in the appropriate units.
m=1.02 × 10^-5

pulled from another response, hope it helps

3 0

3 years ago

Read 2 more answers