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Likurg_2 [28]
3 years ago
6

Identify the two main materials that can be found in the inner and outer core layers of the earth.

Chemistry
1 answer:
Mrrafil [7]3 years ago
6 0
A. Iron D. Nickel hope this helps you!
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MnS+HCl H 2 S+MnCl 2 Which set of coefficients would balance the equation?
Gnom [1K]

Answer:

1,2,1,1

Explanation:

5 0
3 years ago
What best explains why sodium is more likely to react with another element than an element such as neon?
hichkok12 [17]

Answer:

1 fewer electron

Explanation:

5 0
3 years ago
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Ethyl alcohol (ch3ch2oh) is/is not soluble in water. 1. is; all organic molecules are soluble in water. 2. is; ethyl alcohol exh
nata0808 [166]
Answer:
            Ethyl alcohol is soluble in water because <span>ethyl alcohol exhibits dipole-dipole and h-bonding interactions with water.

Explanation:
                   Ethyl alcohol and water are miscible in each other because both are polar in nature and "Like dissolves Like".
                   The bond between oxygen and hydrogen atoms, both in alcohol and water are polar in nature and results in intermolecular hydrogen bond interactions between them as hydrogen bonding results when hydrogen atom in one molecule directly attached to highly electronegative atoms like fluorine, oxygen and nitrogen forms interaction with higly electronegative atom of neighbor atom.</span>
5 0
3 years ago
If I have 12.0 volume of gas at pressure of 1.10 and temperature of 200k, what is the number of moles? (R=8314)
arsen [322]

Answer:

7.94 x 10^6 mol

Explanation:

PV=nRT

n=(PV)/(RT)

n=(1.10*12.0)/(8314*200)

n=7.94 x 10^6 mol

5 0
3 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
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