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Ivan
3 years ago
11

Light from the Sun may look white or yellowish, but it can be separated into a rainbow of colors. True or false?

Chemistry
2 answers:
erik [133]3 years ago
5 0

Yes. It is very trueee...

FrozenT [24]3 years ago
4 0

Answer:

True

Explanation: Different wavelengths get bent in different amounts

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Ants live In colonies
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How many grams of KCO3 are needed to make 20 L of a 2.5 M solution?
PolarNik [594]

Answer:

0.25 L of a solution with a molarity of 6M has 6*0.25 = 1.5 moles of the solute. The molar mass of ammonium sulfate is 132.14 g/mole. The mass of 1.5 moles is 132.14*1.5 = 198.21 g. Therefore 198.12 g of ammonium sulfate are required to make 0.25 L of a solution with a concentration of 6M.

Explanation:

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3 years ago
Precipitation _____. occurs equally all over the globe is the change in state from a gas to a liquid happens when ice changes in
LenKa [72]
<h3><u>Answer;</u></h3>

exceeds evaporation over land

Precipitation<u> exceeds evaporation over land </u>

<h3><u>Explanation;</u></h3>
  • <em><u>In order to maintain earths water balance, evaporation exceeds precipitation over oceans but precipitation exceeds evaporation over land.</u></em>
  • Water evaporates into the atmosphere from the ocean and to a much lesser extent from the continents. Winds transport this moisture-laden air, often great distances, until conditions cause the moisture to condense into clouds and to precipitate and fall.
  • Most precipitation originates by evaporation from the oceans. Over time, water evaporated from the oceans is replenished by inflow of freshwater from rivers and streams.
4 0
3 years ago
1.00 M CaCl2 Density = 1.07 g/mL
Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

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3 years ago
What is the abbreviation for the element with atomic number 11?
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Answer:

Sodium (Na) has atomic number 11.

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