How many neutrons does element X have if its atomic number is 21 and its mass number is 95?

HELPPP! How is a gas different from a solid or a liquid?

Korolek [52]

A gas has more free molecules, and it is air, not a tangible substance. Hope it helps! :)

6 0

3 years ago

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When naming ionic compounds does the roman numeral after the metal cation tell you how many anions will be there?

murzikaleks [220]

Answer:

no

Explanation:

the roman numeral tells you that the metal is a transition metals. transition metals have more than one oxidation state.

8 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

What happens to the total mass of a substance undergoing a physical change

Vlad1618 [11]

Answer: It stays the same

Explanation:

3 0

4 years ago

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What type of claim must be supported by scientifically valid evidence for it to be approved for use on a food label? A. Health c

Lelu [443]

Option A. Health claim

Health claim must be supported by scientifically valid evidence to be approved in using for food label.

This is also a food guide to all consumers the food content implied by manufacturers to their products.

8 0

4 years ago

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