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3 years ago

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Why does the concentration of a product increase as the amount of the reactant decreases? (More 10+ pt questions on profile)

1 answer:

skad [1K]3 years ago

8 0

As reactant concentration decreases, the forward. reaction slows. As product concentration increases, the reverse reaction becomes faster. The forward. reaction will continue to slow and the reverse reaction will continue to increase until they are the same.Then the situation will be at equilibrium.

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What is the volume of stank of nitrogen that contains 17 moles of nitrogen at 34 C under 12,000Pa?

Otrada [13]

Answer:

3626.76dm³

Explanation:

Given parameters:

Number of moles of Nitrogen in tank = 17moles

Temperature of the gas = 34°C

Pressure on the gas = 12000Pa

Unkown:

Volume of the tank, V =?

Converting the parameters to workable units:

We take the temperature from °C to Kelvin

K = 273 + °C = 273 + 34 = 307k

Taking the pressure in Pa to atm:

101325Pa = 1atm

12000Pa = 0.118atm

Solution:

To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.

It is expressed as PV = nRT

The unknown is the Volume and we make it the subject of the formula

V = $\frac{nRT}{P}$

Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹

Therefore V = $\frac{17 x 0.082 x 307 }{0.118}$ = 3626.76dm³

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3 years ago

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Which is not a name to describe global winds A) Polar Easterlies B) Prevailing Westerlies C) Trade Winds D) Tropical Northerlies

SVEN [57.7K]

D Tropical northerlies

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3 years ago

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Im hvaing a hard time getting the right answer

Tanya [424]

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!

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3 years ago

What is the answer to 2.5 * 22.5

nadya68 [22]

Your answer to 2.5*22.56 is 56.25

4 0

3 years ago

What is the percentage by mass of sulphur in Al2(SO4)3[A12SO4 =<br>342g/mol, S = 32]

valentinak56 [21]

Answer: The percentage by mass of sulphur in $Al_2(SO_4)_3$ is 9.36%

Explanation:

Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.

${\text {Mass percentage}}=\frac{\text {mass of sulphur}}{\text {Total mass}}\times 100\%$

Given: mass of sulphur = 32 g/mol

mass of $Al_2(SO_4)_3$ = 342 g/mol

Putting in the values we get:

${\text {Mass percentage}}=\frac{32g/mol}{342g/mol}\times 100\%=9.36\%$

The percentage by mass of sulphur in $Al_2(SO_4)_3$ is 9.36%

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3 years ago