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3 years ago

What is this atom structure?

Chemistry

2 answers:

Alina [70]3 years ago

5 0

Protons, electrons, and neutrons.

Fudgin [204]3 years ago

4 0

Answer:

Lol

Explanation:

You might be interested in

I am doing an experiment that calls for 200g of H2SO4. How many moles of H2SO4 is this?

Mila [183]

Answer: 2.04 moles

Explanation:

Mass of H2SO4 = 200g

RMM of H2SO4 = (2*1) + 32 + (16*4)

= 2 +32 +64 = 98

n = Mass concentration / RMM

n = 200/98

n = 2.04moles

4 0

3 years ago

Who else just copy and pastes the questions from edge?

Kobotan [32]

Answer:

haha

Explanation:

yea because my phat a22 is to lazy

5 0

3 years ago

What is the orbital diagram of Ni^2+?

Crazy boy [7]

Answer is 3d 4s

and I suggest to also download Socratic it is also very helpful

4 0

4 years ago

You are given an unknown mixture containing NaCl and NaHCO3. When you carry out the heating exactly as described in part A, only

SOVA2 [1]

Answer:

1.52g NaHCO3 were in the original mixture.

Mass percent: 64.1%

Explanation:

When NaHCO3 heats it descomposition occurs as follows:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g).

The loss in mass is because of the evaporization of CO2 and H2O. As both are in the same porportion, its molar mass is the sum of both compounds (44g/mol + 18g/mol = 62g/mol)

Loss in mass: 74.80g - 74.24g = 0.56g.

In moles:

0.56g * (1mol / 62g) = 0.00903 moles of gas.

As 1 mole of the gases comes from 2 moles of NaHCO3:

<em>Moles NaHCO3:</em>

0.00903 moles of gas * (2 moles NaHCO3 / 1 mole gas) = 0.018 moles NaHCO3.

In grams (Molar mass NaHCO3: 84g/mol):

0.018 moles NaHCO3 * (84g / mol) = 1.52g NaHCO3 were in the original mixture.

The mass of the mixture was:

74.80g - 72.428g = 2.372g

That means mass percent of NaHCO3 is:

(1.52g / 2.372g) * 100 = 64.1%

7 0

4 years ago

Calculate the number of moles in the following: 2.8 X 10^24 atoms of Cl2

vova2212 [387]

Answer:

<h3>The answer is 4.65 moles</h3>

Explanation:

To find the number of moles given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 2.8 × 10²⁴ atoms of Cl2

So we have

$n = \frac{2.8 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 4.65116279069...$

We have the final answer as

<h3>4.65 moles</h3>

Hope this helps you

7 0

3 years ago