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2 years ago

14

How many moles are there in 1 mole of argon?

1 answer:

disa [49]2 years ago

5 0

the atomic mass of argon is 39.948. That means that one mole of argon weighs 39.948 grams (39.948 g/mol).

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This answer please :))

r-ruslan [8.4K]

Answer: ya this one

Explanation: this is the one

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3 years ago

Both protons and neutrons (and their anti-particles) froze out:

VladimirAG [237]

I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge.

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3 years ago

Please help me with this.

zlopas [31]

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2 years ago

an unknown molecule is found to consist of 24.2% carbon by mass, 4.0% hydrogen by mass and the remaining mass is due to chlorine

Paha777 [63]

Answer:

C3 H6 Cl 3

Explanation:

C -24.2%

H - 4.0%

Cl - (100-24.2 - 4.0)=73.8 %

We can take 100g of the substance, then we have

C -24.2 g

H - 4.0 g

Cl - 73.8 g

Find the moles of these elements

C -24.2 g/12.0 g/mol =2.0 mol

H - 4.0 g/1.0 g/mol = 4. 0 mol

Cl - 73.8 g/ 35.5 g/mol = 2.1 mol

Ratio of these elements gives simplest formula of the substance

C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1

CH2Cl

Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol

Real molar mass = 150 g/mol

real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3

So, Real formula should be C3 H6 Cl 3.

4 0

3 years ago

Read 2 more answers

Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answ

inessss [21]

Answer:

$\Delta S^{0}$ for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: $\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)$

$\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]$

where $S^{0}$ represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

$\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K$

So, $\Delta S^{0}$ for the given reaction is -99.4 J/K

7 0

3 years ago