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Brilliant_brown [7]
3 years ago
15

Hello! I need a bit of help and Im kinda in a bit of a hurry so please help! Thank you :D

Chemistry
2 answers:
Scorpion4ik [409]3 years ago
8 0
The first one is substance 3
The second one is Oxygen, Helium, and carbon dioxide
The third one is the oil floats on top of the water
aalyn [17]3 years ago
8 0

Answer:

so for the first one it is the third one the second one it is carbon dioxide and for the third one,. Gasoline is an oil product. It floats and doesn't mix with water.

Explanation:

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Which material is a mixture 1 water 2 air 3 methane 4 magnesium
amm1812

Answer: Air

Explanation: Pure air is a mixture of several gases that are invisible and odorless. Consists of 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide.

8 0
3 years ago
You are part of a team of scientists and engineers in charge of launching a new spacecraft to venus. What are two ways in which
kap26 [50]

Answer: A barrier should be created to overcome the atmosphere of the Venus, while launching spacecraft to Venus.

Explanation:

The atmosphere of Venus consists of 96.5% carbon dioxide, other composition includes nitrogen and other gases in trace amounts. The large amount of carbon dioxide in the atmosphere can extinguish the missile of the launcher of spacecraft thus it will become difficult in launch of spacecraft to the Venus.

8 0
3 years ago
A solution has a higher boiling point than its associated pure solvent does.
marta [7]

Answer:

4 boiling point elevation

7 0
3 years ago
List and describe the three rock types in the rock cycle including two or more processes involved in the formation of each type
Ludmilka [50]

Answer:    The three main rock types are igneous, metamorphic and sedimentary.

   The three processes that change one rock to another are crystallization, metamorphism, and erosion and sedimentation.

   Any rock can transform into any other rock by passing through one or more of these processes. This creates the rock cycle.

Explanation:

5 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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