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Alona [7]
3 years ago
10

How moles are in 45.7g of CH3?

Chemistry
1 answer:
miskamm [114]3 years ago
6 0

first find the atomic weight of CH3 which would be

atomic weight: 12.011 (3×1.008) = 36.32 g/mol

then find the moles in the given mass

36.32 ÷ 45.7 = 0.794

I HOPE I'M NOT WRONG I HAVENT DONE CHEM IN SO LONG

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Building Vocabulary
Juliette [100K]

Answer:

Explanation:

Building Vocabulary

Match each term with its definition by writing the letter of the correct definition on

the line beside the term in the left column.

5. nucleus   b

6. proton     f

7. neutron   h

8. electron  d

9. atomic number    g

10. isotopes              c

11. mass number      a

12. energy level       e

a. the sum of protons and neutrons in the nucleus of an

atom

b. the very small center core of an atom

c. atoms of the same element that differ in the number

of neutrons, but have the same number of protons

d. the particle of an atom that moves rapidly in the

space outside the nucleus

e. a specific amount of energy related to the movement

of electrons in atoms

f. the particle of an atom with a positive charge

g. the number of protons in the nucleus of every atom

of an element

h. the particle of an atom that is neutral

-. mass number  a.

12. energy level    e

5 0
2 years ago
A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?
Viktor [21]

The answer for the following problem is mentioned below.

  • <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules.</em></u>

Explanation:

Given:

mass of calcium phosphate (Ca_{3}(PO_{4} )_{2} ) = 125.3 grams

We know;

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = 120 + 3(95)

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} )  = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions(N_{A}) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

\frac{m}{M} =\frac{N} }{}N÷N_{A}

\frac{405}{125.3} =\frac{N}{6.023*10^23}

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules</em></u>

3 0
3 years ago
How do i solve it and what is the answer? ​
Masja [62]

Umm there’s no picture

6 0
3 years ago
What are 3 common uses for the Uranium Element?
ipn [44]
Uranium provides nuclear fuel used generate electricity in nuclear power station,also used by the military to power nuclear submarines and in nuclear weapons.
4 0
3 years ago
What kind of change is heating of mercuric oxide? ​
mario62 [17]

Answer:

Chemical change.

Explanation:

When it is heated it decomposes into mercury and oxygen gas. The mercury oxide reactant becomes the silver color of mercury. Hence, a color change can be noticed throughout the reaction.

7 0
3 years ago
Read 2 more answers
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