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2 years ago

20 PTS!!! What is the volume of a gas at 90.0°C, if it occupies 1.41 L at 170°C?

Chemistry

2 answers:

Fofino [41]2 years ago

5 0

Answer:

The new volume of the gas is 1.72L

Explanation:

V1 = 1.41L

V2 = ?

T1 = 90°C = (90+273.15)K = 363.15K

T2 = 170°C =(170+273.15)K = 443.15K

To find the new volume V2 of the gas, we apply Charles law which states that the volume of a fixed mass of gas is directly proportional to its temperature provided that pressure remains constant.

V = KT, K = V1 / T1 = V2 /T2 = V3 /T3.......= Vn / Tn

V1 / T1 = V2 / T2

Solve for V2,

V2 = (V1 * T2) / T1

V2 = (1.41 * 443.15) / 363.15

V2 = 1.72L

The new volume of the gas is 1.72L

Margarita [4]2 years ago

3 0

Answer:

1.72 L

Explanation:

Initial Temperature of the gas T =90°+273= 363 K

Initial Volume of the gas V1= 1.41 L

Final temperature of the gas T2= 170°c +273= 443 K

Final volume of the gas V2= ????

Using Charles law;

V1/T1 = V2/T2

V1T2= V2T1

V2= V1T2/T1

V2= 1.41×443/363

V2= 1.72 L

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What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?

tankabanditka [31]

The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M

3 0

2 years ago

Read 2 more answers

11. A sample contains 25% water and weighs 201 grams. Determine the grams of water in the

77julia77 [94]

Answer:

$m_{water}=50.25g$

Explanation:

Hello,

In this case, considering that the by-mass percent of water is:

$\% m/m=\frac{m_{water}}{m_{sample}}*100\%$

Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:

$m_{water}=\frac{\%m/m*m_{sample}}{100\%}\\ \\m_{water}=\frac{25\%*201g}{100\%}\\ \\m_{water}=50.25g$

Best regards.

5 0

2 years ago

Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee

lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a) 130 g of C7H14O2 ---------------- 1 mol of C7H14O2

1 x 10 ⁻⁶ g --------------- x

x = 7.7 x 10 ⁻⁹ mol

1 mol of C7H14O2 -------------- 6 .023 x 10 ²³ molecules

7.7 x 10⁻⁹ mol -------------- x

x = 4.64 x 10¹⁵ molecules

b) 130 g of C7H14O2 ---------------- 1 mol of C7H14O2

1 x 10⁻⁶ C7H14O2 ----------------- x

x = 7.7 x 10 ⁻⁹ mol of C7H14O2

1 mol of C7H14O2 --------------- 2 mol of O2

7.7 x 10 ⁻⁹ ---------------- x

x = 1.54 x 10⁻⁸ mol of O2

1 mol of O2 ----------------- 6.023 x 10 ²³ atoms

1.54 x 10 ⁻⁸ ---------------- x

x = 9.28 x 10 ¹⁵ atoms of O2

8 0

2 years ago

in sample of elemental bromine, 55% or the atoms are Br-79, and the remainder are Br-81. if this sample is typical of naturally

Gwar [14]

Answer:

The average atomic mass of bromine is 79.9 amu.

Explanation:

Given data:

Percentage of Br⁷⁹ = 55%

Percentage of Br⁸¹ = 45%

Average atomic mass of bromine = ?

Formula:

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

Average atomic mass = [55 × 79] + [81 ×45] / 100

Average atomic mass = 4345 + 3645 / 100

Average atomic mass = 7990 / 100

Average atomic mass = 79.9 amu

The average atomic mass of bromine is 79.9 amu.

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2 years ago

Which of these facts best illustrates why regulation of alcohol consumption is necessary

exis [7]

So that people don’t break laws or drive under the influence.

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2 years ago