All categories

3 years ago

How many grams are in 2.30 x 10^24 atoms in silver

Chemistry

1 answer:

PtichkaEL [24]3 years ago

4 0

411.984 grams

The atomic mass of Silver is 107.8682 by the periodic chart I have here. That means that 1 mole of Silver masses

107.8682 grams if the Silver is pure.

I mole of any element like silver is also 6.022 X 10^23 atoms of silver.

10^24 atoms is therefore more than a mole. You specified

2.3 x 10^24 atoms, so the number of moles will be

(2.3 X 10^24)divided by 6.022 X 10^23 = 3.819 moles

this number X 107.8682 grams/mole gives us 411.984 grams

You might be interested in

A) An iron nail is attracted to a magnet.

11111nata11111 [884]

there's no question on here

4 0

3 years ago

Read 2 more answers

Which statement describes a change that occurs during a chemical reaction? A. Atoms in the original substances are changed into

trapecia [35]

Answer:

B. Atoms in the original substances are arranged in a different way to make new substances.

Explanation:

The best statement that describes a change that occurs in a chemical reaction is that atoms in the original substances are arranged in a different way to make new substances.

Chemical reactions obey the law of conservation of matter.
The law postulates that "matter is neither created nor destroyed in a chemical reaction but they are simply rearranged".
Therefore, atoms of compounds forms new bonds by rearranging to give a new product.

6 0

3 years ago

Read 2 more answers

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

aalyn [17]

Answer:

2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, $Me_{2}S$ etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Reaction steps are shown below.

8 0

3 years ago

How many moles are there in 3.612 x 1024 molecules of CaO?

melamori03 [73]

The number of moles in 3.612 x 10²⁴ molecules of CaO is 6 moles.

<h3>Number of moles in the molecules</h3>

The number of moles in 3.612 x 10²⁴ molecules of CaO is calculated as follows;

6.02 x 10²³ molecules = 1 mole

3.612 x 10²⁴ molecules = ?

= (3.612 x 10²⁴ ) / (6.02 x 10²³ )

= 6 moles

Thus, the number of moles in 3.612 x 10²⁴ molecules of CaO is 6 moles.

Learn more about number of moles here: brainly.com/question/15356425

3 0

2 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago