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MrRa [10]
3 years ago
12

What is the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3? (5 points) Group of answer choices 203 grams 161 gr

ams 214 grams 187 grams
Chemistry
1 answer:
Nitella [24]3 years ago
8 0

Answer:

203 grams

Explanation:

The no. of moles of (6.3 x 10²⁴ molecules--Avagadros number) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.

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The atomic number is determined only by the number of ________ in the nucleus of an atom. But, in a neutral atom it also represe
sveticcg [70]

Hello!

The atomic number is determined only by the number of protons in the nucleus of an atom. But, in a neutral atom it also represents the number of electrons in the electron cloud.


Neutrons are only important in the nucleus for helping us find atomic weight, which varies as we move along the perodic table and does not always equal the same amount of it's atomic number. Which is why it would not be a suitable answer for the first blank space. Electrons do not work either as they do not exist inside the nucleus but rather outside the atom.

The second space, since it states is in the electron cloud, we can deduct that electrons would be an appropriate answer there.

If you need anymore help feel free to ask, but I hope this answers your question.

5 0
2 years ago
Does anyone have the student exploration sheet answers for the Drug Dosage (Forensic Science) Gizmos lab?​
Novay_Z [31]

Answer:

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8 0
3 years ago
The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample
IgorLugansk [536]

Answer:

6.2 calories

Explanation:

Data Given:

change in temperature = 20 °C

specific heat of gold = 0.031 calories/gram °C

mass of gold = 10.0 grams

Amount of Heat = ?

Solution:

Formula used

             Q = Cs.m.ΔT

Where:

Q = amount of heat

Cs = specific heat of gold = 0.031 calories/gram °C

m = mass

ΔT = Change in temperature

Put values in above equation

                Q = 0.031 calories/gram °C x 10.0 g x 20 °C

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So option A is correct = 6.2 calories

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Explanation:

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lina2011 [118]

Explanation:

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Hoped this helped, 2Trash4U

5 0
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