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Hunter-Best [27]
4 years ago
10

Consider a sample of helium and a sample of neon, both at 30.0°C and 1.5 atm. Both samples have a volume of 5.0 liters. Which st

atement concerning these samples is not true? Consider a sample of helium and a sample of neon, both at 30.0°C and 1.5 atm. Both samples have a volume of 5.0 liters. Which statement concerning these samples is not true?
A) Each sample weighs the same amount.
B) The density of the neon is greater than the density of the helium.
C) Each sample contains the same number of atoms of gas.
D) Each sample contains the same number of moles of gas.
E) none of the above
Chemistry
1 answer:
Triss [41]4 years ago
7 0

Answer:

A.

Explanation:

Using the ideal gas equation, we can calculate the number of moles present. I.e

PV = nRT

Since all the parameters are equal for both gases, we can simply deduce that both has the same number of moles of gases.

The relationship between the mass of each sample and the number of moles can be seen in the relation below :

mass in grammes = molar mass in g/mol × number of moles.

Now , we have established that both have the same number of moles. For them to have the same mass, they must have the same molar masses which is not possible.

Hence option A is wrong

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Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.

5 0
3 years ago
I really need help with this. Be serious. If you answer something random just for the points I will report you to brainly.
sashaice [31]

Answer:

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Explanation:

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8 0
2 years ago
What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C
Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = \frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\

Now use the molar ratio to find how many moles of NaOH would be required to neutralize 5.5537*10^{-3} mol\\ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

5.5537*10^{-3} mol\\ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L

5 0
4 years ago
Name the two molecules Nн нннн
arlik [135]

Answer:

1: C5H12

2:C5H11

Explanation:

nakqkchlqosnx

4 0
3 years ago
Balanced equation for the neutralization of these 2. HCl , Ca(OH)2
Ghella [55]
Balanced equation:

<span>2 HCl + 1 Ca(OH)</span>₂<span> = 1 CaCl</span>₂<span> + 2 H</span>₂<span>O
</span>
hope this helps!


5 0
3 years ago
Read 2 more answers
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