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Mamont248 [21]
2 years ago
5

# of protons = 48 Cadmium Titanium Oxygen Krypton

Chemistry
1 answer:
Dafna11 [192]2 years ago
7 0

Answer:

Cadmium +p= 48

Titanium p=22

Oxygen p=8

Krypton p=36

So the answer is Cadmium which has #48 protons

Explanation:

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A student investigated the factors that affect math grades. Which statement from the lab report best represents a conclusion for
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Students who spend less time studying after school get lower math grade

Option D is the correct answer

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Mathematics is one of the most important aspect of knowledge in the education.

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Students who spend less time studying after school get lower math grade

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1 year ago
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crimeas [40]

Answer:

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Explanation:

3 0
3 years ago
The number of proton, neutrons and electrons in oxygen atom
Ksenya-84 [330]

Answer:

in an oxygen atom there are:

protons:8

electrons:8

neutrons:8

Explanation:

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4 0
2 years ago
Can someone help please I will give Brainlyiest
kotykmax [81]

Answer:

See Explanation

Explanation:

moles of NH₃ = 11.9g/17.03 g/mol = 0.699 mole

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Theoretical yield of CN₂OH₄ = (0.349 mole)(60 g/mole) = 20.963 grams

%Yield = Actual Yield/Theoretical Yield x 100%

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7 0
3 years ago
Read 2 more answers
When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced
Vika [28.1K]

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO_2 and H_2O

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

When only 2.5 mol of O_2 are consumed in order to complete the reaction, ________ mol of CO_2 are produced.

Answer: Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

Explanation:

The balanced chemical equation is:

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

According to stoichiometry :

5 moles of O_2 produce = 3 moles of  CO_2

Thus 2.5 moles of O_2 will produce = \frac{3}{5}\times 2.5=1.5 moles of  CO_2

Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

4 0
2 years ago
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