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Juli2301 [7.4K]
3 years ago
5

How many molecules are there in 24 grams of FeF3?

Chemistry
1 answer:
goldfiish [28.3K]3 years ago
7 0
24 gFeF3 x (1 mol FeF3/grams FeF3) x (6.02x10^23 molecules FeF3/ 1 mol FeF3) Just Calculate Molar Mass of FeF3 and plug into equation
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Why do we use scientific notation or scientific exponential numbers in math calculations?
seraphim [82]
I believe the answer is: in order not to write very big or very small number values
6 0
3 years ago
A sample of Copper absorbs 43.6 KJ of heat, resulting in a temperature rise of 75.0 oC, determine the mass (in Kg) of the copper
yanalaym [24]

Explanation:

here is the answer to your question

3 0
2 years ago
The human body excretes nitrogen in the form of urea, NH₂CONH₂. The key step in its biochemical formation is the reaction of wat
Misha Larkins [42]

Mass perventage of nitrogen in urea, arginine, ornithine is   46.67%, 32.2%,21% respectively.

What is Mass perventage ?

Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100.

Molecular mass of urea (NH2CONH2) is= 2× (Atomic mass of N ) + 4 × (Atomic mass of H) + (Atomic mass of C) (Atomic mass of O) = 60

Mass percentage of N = Total mass of N atoms in the compoundMass of compound×100=2860×100=46.67%

chemical formula for arginine is C6H14N4O2

molar mass of C6H14N4O2=174g/mol

moles of N atoms in C6H14N4O2=4

mass of N atoms=14*4=56g

mass percent of N in C6H14N4O2=(56/174)*100=32.2%

the chemical formula for ornithine is C5H12N2O2

molar mass of C5H12N2O2=132g/mol

moles of N atoms in C5H12N2O2=2

mass of N atoms in C5H12N2O2=2*14=28

mass percent of N in C5H12N2O2=(28/132)*100=21%

To learn more about urea click on the link below:

brainly.com/question/17812875

#SPJ4

7 0
1 year ago
PLZZ!!!- I NEED THIS RIGHT NOWA block is pulled 0.90 m to the right in 2.5 s.
UNO [17]
The average speed is 0.28
4 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
3 years ago
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