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BaLLatris [955]
3 years ago
10

4K2CO3 Subscript for K: Subscript for C: Subscript for O:

Chemistry
1 answer:
katovenus [111]3 years ago
4 0

Answer:

bjbjbjbjbhjhgytuvbgtbgygutf

Explanation:

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Does acid rain make new substances? Construct an argument that uses data from this investigation as evidence for your claim.
Yuri [45]

Answer:

Acid rain, like all acids, generates dissolution of chemical compounds, and reacts together with bases to be able to carry out oxide reduction reactions, which by deduction is very likely to form new substances as a result of the dissolution or acid erosion caused.

Explanation:

Acid rain is more likely to occur in large cities or large sources of pollution, since the excess of carbon dioxide in the atmosphere causes it to increase its partial pressure in a condensed way in the clouds, this is how then this cloud when being loaded with water and then generating the rain drags these masses of condensed carbon dioxide in the form of acid rain.

The degree of acidity is directly proportional to the amount of partial pressure of carbon dioxide in the atmosphere.

5 0
3 years ago
Show your work including units in your answer.<br> Calculate the molar mass of Li4OH
lorasvet [3.4K]
Empirical formula: Li4OH
Answer:
1 mole: 44.771 g
1 gram = 0.022 mole

Explanation:

Element: Li
Percentage by mass: 62.01%
Number of atoms: 4
Mass of atom;: 6.941

Element: O
Percentage by mass: 35.74%
Number of atoms: 1
Mass of atom: 15.9994

Element: H
Percentage by mass: 2.25%
Number of atoms: 1
Mass of atom: 1.00794

4 0
3 years ago
Please help me with this question it's due in like 3hours from now
saw5 [17]

237kj mole for both.....




4 0
3 years ago
Which of the following gases effuses the fastest at a given temperature?
Sholpan [36]

Answer: -

H₂ will diffuse the fastest.

Explanation: -

According to Graham's Law of Diffusion

The rate of diffusion is inversely proportional to the square root of it's density or molar mass. So the lower the molar mass faster the rate of diffusion.

Molar mass of Ne = 20 g / mol

Molar mass of CH₄ = 12 x 1 + 1 x 4 = 16 g /mol

Molar mass of Ar = 40g / mol

Molar mass of H₂ = 1 x 2 = 2 g / mol

Thus H₂ will diffuse the fastest.

7 0
3 years ago
Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to
Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density \rho_{avg}; we have:

\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }

The average atomic weight is:

A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

C_{Fe} = 100 - C_v

From a unit cell volume V_c

V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}

where;

N_A = number of Avogadro constant.

SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

C_{Fe} with 100-C_v

Then:

a^3 = \dfrac   { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) }    {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

Replacing the values; we have:

(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol)   }   }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3)   } }

2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} }  \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }

2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})

\mathbf{C_v = 9.1 \ wt\%}

4 0
3 years ago
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