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Jlenok [28]
3 years ago
11

HELP PLEASE!!!!! show work

Chemistry
1 answer:
muminat3 years ago
5 0
I’m not sure how to show the work
You should download the app chem balance
It’s great for this kind of homework

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Water vapor forming droplets that form clouds directly involve which process
Vinvika [58]
It is formed when water vapor condenses.
5 0
3 years ago
0.68 moles of calcium to calcium atoms​
Oksanka [162]

0.68 \: moles \: calcium \times  \frac{6.02 \times  {10}^{23} \: calcium \: atoms }{1 \: mole \: calcium} \\

The (( mole calcium ))s are simplified so the answer is based on the number of calcium atoms.

0.68 \times 6.02 \times  {10}^{23} = 68 \times  {10}^{ - 2} \times 602 \times  {10}^{ - 2} \times  {10}^{23} \\

= 68 \times 602 \times  {10}^{ - 4} \times  {10}^{23} = 40936 \times  {10}^{19} \\

calcium \: atoms = 40936 \times  {10}^{19}  \\

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

3 0
3 years ago
Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP; is needed to comple
spayn [35]
The balanced chemical reaction would be expressed as follows:

2Na + Cl2 = 2NaCl

We are given the amount of sodium metal to be used reaction. This would be the starting value for the calculations. We do as follows:

6.25 g ( 1 mol / 22.99 g ) ( 1 mol Cl2 / 2 mol Na ) ( 22.4 L / 1 mol ) = 3.045 L Cl2 needed

Hope this answers the question.

6 0
3 years ago
I have a mixture of sand salt and water from the sea. Describe in detail how I separate these three substances and explain why e
Ivanshal [37]

Answer:

You can separate a mixture of sand, salt and water by decantation and evaporation. Decantation will separate sand from salt and water. Sand remains in the original vessel, whereas salt and water is transfered to new vessel. Evaporation will separate salt from water.

3 0
2 years ago
The PH solution of NH3 of 0.950molar solution is 11.612. find the Kb​
Vaselesa [24]

Answer:

1.8 x 10⁻⁵

Explanation:

 NH3(aq)  +  H2O(l)  ⇄ NH4⁺(aq)  +   OH⁻(aq)

I  0.95                              0                    0

C -x                                 +x                  +x

E 0.95-x                           x                    x

Kb= [NH₄⁺] [OH⁻] / (  NH₃) = x²/ (0.95-x )

P(OH) = 14-PH = 14-11.612 = 2.388

(OH)⁻¹ = 10⁻²°³⁸⁸ = 4.09 x 10⁻³ = x

Kb = (4.09 x 10⁻³)²/ (0.95-4.09 x 10⁻³)

= 1.8 x 10⁻⁵

8 0
3 years ago
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