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melisa1 [442]
3 years ago
7

What is the percentage composition of water in crystallization of oxalic acid​

Chemistry
1 answer:
docker41 [41]3 years ago
6 0

Answer:

5.17 percentage is water

Explanation:

answer from gauth math

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How many moles of CaCl2 are contained in 0.448 L of a 0.85 M CaCl2 solution?
kati45 [8]

Answer:

0.3808

Explanation:

number of moles,n=Conc.XVol.

hence 0.85X0.448

3 0
3 years ago
2H2+ O2 —> 2H2 O
lukranit [14]

Answer:bubuvuvuvftctcrchj

Explanation:

8 0
3 years ago
How is the mass number related to the number of protons and neutrons an atom has
matrenka [14]
Mass is equal to protons plus neutrons
To find the protons its the atomic number
8 0
3 years ago
Magsulat sa iyong sailing lyrics na kanta the tune of meron leron sinta<br>​
xxMikexx [17]

Answer:

I am sailing

I am sailing

Home again

'Cross the sea

I am sailing

Stormy waters

To be near you

To be free

I am flying

I am flying

Like a bird

'Cross the sky

I am flying

Passing high clouds

To be with you

To be free

Can you hear me, can you hear me

Through the dark night, far away

I am dying, forever crying

To be with you, who can say

Can you hear me, can you hear me

Through the dark night far away

I am dying, forever crying

To be with you, who can say

We are sailing, we are sailing

Home again

'Cross the sea

We are sailing

Stormy waters

To be near you

To be free

Oh Lord, to be near you, to be free

Oh Lord, to be near you, to be free

Oh Lord, to be near you, to be free

Oh Lord

Explanation:

I hope this is the right one

3 0
3 years ago
A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume
alexandr402 [8]

Answer:

x_{N_2F_2}= 0.415\\\\x_{SF_6}=0.585

Explanation:

Hello!

In this case, since the mole fraction of both gases in the tank is computed via:

x_{N_2F_2}=\frac{n_{N_2F_2}}{n_{N_2F_2}+n_{SF_6}} \\\\x_{SF_6}=\frac{n_{SF_6}}{n_{N_2F_2}+n_{SF_6}}

It means we need to compute the moles of each gas, just as it is shown down below:

n_{N_2F_2}}=5.53gN_2F_2*\frac{1molN_2F_2}{66.01gN_2F_2} =0.0838molN_2F_2\\\\n_{SF_6}=17.3gSF_6*\frac{1molSF_6}{146.06gSF_6} =0.118molSF_6

Thus, the mole fractions turn out:

x_{N_2F_2}=\frac{0.0838mol}{0.0838mol+0.118mol}= 0.415\\\\x_{SF_6}=\frac{0.0838mol}{0.0838mol+0.0838mol}=0.585

Best regards!

8 0
3 years ago
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