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4 years ago

If you have 3.01 E24 molecules of CO, how many moles of CO do you have?

Chemistry

1 answer:

MA_775_DIABLO [31]4 years ago

5 0

Answer: 5moles

Explanation:

1mole of a substance contains 6.02x10^23 molecules.

Then, 1mole of CO also contains 6.02x10^23 molecules.

If 1 mole of CO2 contains 6.02x10^23 molecules, it means Xmol of CO contains 3.01 E24 ie 3.01x10^24 molecules

Xmol of CO = 3.01x10^24 / 6.02x10^23

Xmol = 5moles

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Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer t

nekit [7.7K]

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

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

<u>Where:</u>

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

$n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol$

The enthalpy change is:

$\Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ$

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

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

A ____________ is a large body that moves around a star?

Travka [436]

Answer:

A Planet

Explanation:

The earth for example, is a large body that orbits the sun, our local star

4 0

3 years ago

Read 2 more answers

PLEASE PLEASE PLEASE HELP! WILL AWARD BRAINLIESTPLEASE ANSWER AND STOP JUST LOOKING AT ITTTTT

Goshia [24]

Answer:

answer - c

answer - a

Explanation:

<h2>I hope answer correct</h2><h2><em><u>pl</u></em><em><u>ease</u></em><em><u> like</u></em><em><u> me</u></em></h2>

3 0

3 years ago

Calculate the molarity (M) of a solution containing 49.0 grams of H3PO4 in 500 mL of solution.

vlada-n [284]

Answer

The molarity (M) of the H3PO4 solution = 1.0 M

Explanation

Given:

Mass of H3PO4 = 49.0 grams

Volume of the solution = 500 mL = 500/1000 = 0.5 L

What to find:

The molarity (M) of the H3PO4 solution.

Step-by-step solution:

Step 1: Convert 49.0 grams H3PO4 to moles using the mole formula.

$Mole=\frac{Mass}{Molar\text{ }mass}$

The molar mass of H3PO4 = 97.994 g/mol

So,

$Mole=\frac{49.0\text{ }g}{97.994\text{ }g\text{/}mol}=0.50\text{ }mol$

Step 2: Calculate the molarity of the solution using the molarity formula.

$Molarity=\frac{Mole}{Volume\text{ }in\text{ }L}$

Putting mole = 0.50 mol and volume = 0.50L into the formula, we have;

$Molarity=\frac{0.50mol}{0.50L}=1.0\text{ }M$

The molarity (M) of the H3PO4 solution = 1.0 M

5 0

1 year ago