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Lemur [1.5K]
3 years ago
6

Describe how you think the tiny dust particles acquired a charge.

Chemistry
1 answer:
miskamm [114]3 years ago
7 0
Whenever they move, they get a charge from the moment. <span />
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PlEasE help ASAP <br> Snskskssmskssm
salantis [7]

Answer:

message me on messenger johnpatrick so I can help you

5 0
3 years ago
If 4.59 g of potassium reacts with 3.6 g of sulfur according to the following reaction, how many grams of potassium sulfide can
EleoNora [17]
First convert the amount of grams you have of each substance to moles. Find your limiting reactant by calculating how many grams are needed to complete this reaction. If done correctly, you would see that we need .226 moles of Potassium to complete this reaction. However, we only have .118 moles of Potassium, so K must be our limiting reactant. Then use the moles of K to find out how many moles of K^2S are made. Then convert the amount of moles of K^2S to grams and you should get 10.3 g K^2S
8 0
3 years ago
a solution of HCI contains 36 percent HCI,by mass and calculate the mole fraction of HCI in the solution?​
denis-greek [22]

Explanation:

You have a solution that contains 36 g HCl dissolved in 64 g water

Molar mass HCl = 36.45 g/mol

Mol HCl in 36 g = 36 g / 36.45 g/mol = 0.9876 mol

Molar mass H2O = 18 g/mol

Mol H2O in 64 g = 64 g / 18 g/mol = 3.5556 mol

Total mol = 0.9875 + 3.5556 = 4.5431 mol

Mol fraction HCl = 0.9876 mol / 4.5431 mol = 0.2174

Mol fraction H2O = 3.5556 / 4.5431 = 0.7826

The answer should have 2 significant digits:

Mol fraction HCl = 0.22

Mol fraction H2O = 0.78

Mol fraction has no units.

THAT IS HELPFUL FOR YOU

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4 0
2 years ago
9. At equilibrium a 2 L vessel contains 0.360
klio [65]

Answer:

Ke = 34570.707

Explanation:

  • H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

3 0
3 years ago
Which of the elements shown will not form ions, and why will they not do so?
grigory [225]

No elements visible!

Ions form between metals and non-metals.

Hope this helps!

6 0
2 years ago
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