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3 years ago

How many moles of solute are in 53.1 mL of 12.5M HCI?

1 answer:

6 0

Molarity = moles of solute/volume of solution in liters.

From this relation, we can figure out the number of moles of solute by multiplying the molarity of the solution by the volume in liters.

We have 53.1 mL, or 0.0531 L, of a 12.5 M, or 12.5 mol/L, solution. Multiplying 12.5 mol/L by 0.0531 L, we obtain 0.664 moles. So, in this volume of solution, there are 0.664 moles of solute (HCl).

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A chemist has a 12.5 liter sample of neon gas at 1.00 atm and 30 c. If the chemist compresses the sample to 10.5 liters and hold

ycow [4]

Answer:

254.5 K

Explanation:

Data Given

initial volume V1 of neon gas = 12.5 L

final Volume V2 of neon gas = 10.5 L

initial Temperature T1 of neon gas = 30 °C

convert Temperature to Kelvin

T1 = °C +273

T1 = 30°C + 273 = 303 K

final Temperature T2 of neon gas = ?

Solution:

This problem will be solved by using Charles' law equation at constant pressure.

The formula used

V1 / T1 = V2 / T2

As we have to find out Temperature, so rearrange the above equation

T2 = V2 x T1 / V1

Put value from the data given

T2 = 10.5 L x 303 K / 12.5 L

T2 = 254.5K

So the final Temperature of neon gas = 254.5 K

8 0

3 years ago

How many grams of 2.5% "h" cream should be mixed with360 grams of 0.25% "h" cream to make a 1% "h" cream?

Leona [35]

Answer is: 180 <span>grams of 2.5% "h" cream should be mixed.
</span>ω₁ = 2.5% ÷ 100% = 0.025.
ω₂ = 0.25% ÷ 100% = 0.0025.
ω₃ = 1% ÷ 100% = 0.01.
m₂ = 360 g.
m₃ = m₁ + 360 g.
m₁ = ?.
ω₁ · m₁ + ω₂ · m₂ = ω₃ · m₃.
0.025 · m₁ + 0.0025 · 360 g = 0.01 · (m₁ + 360 g).
0.025 · m₁ + 0.9 g = 0.01 · m₁ + 3.6 g.
0.015 · m₁ = 2.7 g.
m₁ = 180 g.

4 0

3 years ago

How long is a grain of rice? Select the best estimate.

grigory [225]

Answer:

0.27 to 0.35 inch i found one that was 0.47 inches Lol

5 0

3 years ago

Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a sedative and hypnotic.

Dmitrij [34]

Answer :

(a) The molar mass of $C_2H_3Cl_3O_2$ is, 165.5 g/mole

(b) The moles of $C_2H_3Cl_3O_2$ is, 3.02 moles

(c) The mass in grams of $2.0\times 10^{-2}$ mole chloral hydrate is, 3.31 g

(d) The number of chlorine atoms in 5.0 g chloral hydrate is, $5.4\times 10^{22}$

(e) The mass of chloral hydrate will be, 1.55 g

(f) The mass of exactly 500 molecules of chloral hydrate is, $1.99\times 10^{23}$

Explanation :

(a) To calculate the molar mass of chloral hydrate.

The formula of chloral hydrate is, $C_2H_3Cl_3O_2$

Atomic mass of carbon = 12 g/mole

Atomic mass of hydrogen = 1 g/mole

Atomic mass of oxygen = 16 g/mole

Atomic mass of chlorine = 35.5 g/mole

Now we have to determine the molar mass of chloral hydrate.

$\text{Molar mass of }C_2H_3Cl_3O_2$ =2(12g/mole)+3(1g/mole)+3(35.5g/mole)+2(16g/mole)=165.5g/mole[/tex]

The molar mass of $C_2H_3Cl_3O_2$ is, 165.5 g/mole

(b) Now we have to determine the moles of $C_2H_3Cl_3O_2$ .

$\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{500.0g}{165.5g/mole}=3.02moles$

The moles of $C_2H_3Cl_3O_2$ is, 3.02 moles

(c) Now we have to determine the mass in grams of $2.0\times 10^{-2}$ mole chloral hydrate.

$\text{Mass of }C_2H_3Cl_3O_2=\text{Moles of }C_2H_3Cl_3O_2\times \text{Molar mass of }C_2H_3Cl_3O_2$

$\text{Mass of }C_2H_3Cl_3O_2=(2.0\times 10^{-2}mole)\times (165.5g/mole)=3.31g$

The mass in grams of $2.0\times 10^{-2}$ mole chloral hydrate is, 3.31 g

(d) To calculate the number of chlorine atoms are in 5.0 g chloral hydrate.

First we have to determine the moles of $C_2H_3Cl_3O_2$ .

$\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{5g}{165.5g/mole}=0.03moles$

Now we have to calculate the number of chlorine atoms in chloral hydrate.

In $C_2H_3Cl_3O_2$ , there are, 2 carbon atoms, 3 hydrogen atoms, 3 chlorine atoms and 2 oxygen atoms.

As, 1 mole of $C_2H_3Cl_3O_2$ contains $3\times 6.022\times 10^{23}$ chlorine atoms

So, 0.03 mole of $C_2H_3Cl_3O_2$ contains $0.03\times 3\times 6.022\times 10^{23}=5.4\times 10^{22}$ chlorine atoms

The number of chlorine atoms in 5.0 g chloral hydrate is, $5.4\times 10^{22}$

(e) To calculate the mass of chloral hydrate would contain 1.0 g Cl.

As, $3\times 35.5g$ of chlorine present in 165.5 g of $C_2H_3Cl_3O_2$

So, 1 g of chlorine present in $\frac{165.5}{3\times 35.5}=1.55g$ of $C_2H_3Cl_3O_2$

The mass of chloral hydrate will be, 1.55 g

(f) To calculate the mass of exactly 500 molecules of chloral hydrate.

As, $6.022\times 10^{23}$ molecules of chloral hydrate has 165.5 g mass of chloral hydrate

So, 500 molecules of chloral hydrate has $\frac{6.022\times 10^{23}}{500}\times 165.5=1.99\times 10^{23}$ mass of chloral hydrate

The mass of exactly 500 molecules of chloral hydrate is, $1.99\times 10^{23}$

3 0

3 years ago

August kekule described the various ring structures of the carbon compound benzene. What type of chemist would he be considered

Andreyy89

B. He would be considered an Organic Chemist since Organic Chemistry is the study of Carbon and its compounds.

3 0

3 years ago

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