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3 years ago

Which of the following observations would indicate that a material is a pure substance

Chemistry

2 answers:

Lena [83]3 years ago

3 0

What is the following

yaroslaw [1]3 years ago

3 0

It does not dissolve in water.

It changes color after heating.

It cannot be physically broken down into different types of atoms.

It can be physically broken down into different types of molecules.

^the options

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Combustion of hydrocarbons such as undecane (_____) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's

____ [38]

Answer: Undecane, C11H24 is a liquid hydrocarbon from the alkane homologue ( family ).

Explanation:

All hydrocarbons burn in air to release H20 and CO2.

Combustion of hydrocarbons thus increases the percentage of CO2 present in the atmosphere which is not so good as CO2 is a greenhouse gas. Greenhouse gases produce the greenhouse effect. The green house effect is the warming effect produced when gases present in the atmosphere allow and retain the radiation from the sun to the earth. Examples of greenhouse gases are CO2, CH4, SO3, NO, CFCs. The greenhouse effect leads to an increase in the temperature of the earth subsequently leading to (I) melting of large amounts of ice at the poles causing flooding. (II) rapid evaporation of water from Earth’s surface. (III) decrease in crop yield.

Also, high concentration CFCs ( chlorofluorocarbons ), as widely used in aerosols, coolants and packaging foams, in the atmosphere depletes the ozone layer leading to global warming. In 1996 an international agreement was reached for a complete ban of CFCs and more countries are now going Green ( using alternative energy sources ) so as to minimize emissions of these greenhouse gases

5 0

4 years ago

Which type of reactions form salts?

s344n2d4d5 [400]

Neutralization reactions are the reactions type which form salts.

Explanation:

Salts are formed by ionic bonds when the oxidation states of anions and cations are equal and have opposite signs. So one should be highly electronegative in nature and another should be highly electropositive in nature. So the electropositive element will be ready to give electrons and the electronegative element will be ready to accept all the electrons given by the electropositive element. As a whole the compound will be neutrally charged by adding of equal number of positively charged and negatively charged ions.

The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.

So the salt formation is based on neutralization reactions.

8 0

3 years ago

Which ion in the ground state has the same electron configuration as an atom of neon in the ground state?

Ainat [17]

Cl-

Lets see how

Z for Cl is 9
One electron is added so new Z=10

Electronic configuration

1s²2s²2p⁶

[Ne]

option B is correct

6 0

2 years ago

Read 2 more answers

If 8.74 g of CuNO3 is dissolved in water to make a 0.700 M solution, what is the volume of the solution

Bess [88]

Hope this helps you.

4 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