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4 years ago

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron?

2 answers:

7 0

Energy would be needed to cause an electron to be a valence electron.

Mass cannot be created nor destroyed as the law of conservation of mass states.

trapecia [35]4 years ago

6 0

Answer:energy

Explanation:

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What is the mass of 100 mL of corn oil?

prisoha [69]

Answer:

m=(0.92g/ml)(100ml)

=92g

hope this helps

5 0

3 years ago

How many molecules of water are there in 10 lbs of water?

Scilla [17]

55.556

Explanation:

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3 0

3 years ago

The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?

kodGreya [7K]

Answer : The concentration of Si in kilograms is, $19.55kg/m^3$

Explanation :

As we are given that, the concentration of Si in an Fe-Si alloy is 0.25 wt% that means:

Weight of Si = 0.25 g = 0.00025 kg

Weight of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = $2.32g/cm^3=2.32\times 10^6g/m^3$

Density of Fe = $7.87g/cm^3=7.87\times 10^6g/m^3$

Now we have to calculate the concentration in kilograms of Si per cubic meter of alloy.

Concentration of Si in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}$

Concentration of Si in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}$

Now put all the given values in this expression, we get:

Concentration of Si in kilograms = $\frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}$

Concentration of Si in kilograms = $19.55kg/m^3$

Thus, the concentration of Si in kilograms is, $19.55kg/m^3$

5 0

4 years ago

A certain liquid X has a normal boiling point of 118.90 °C and a boiling point elevation constant Kb = 0.82 °C*kg*mol^-1. Calcul

Fantom [35]

<u>Answer:</u> The boiling point of solution is $1.2\times 10^2^oC$

<u>Explanation:</u>

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\Delta T_b=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

Boiling point of pure liquid = 118.90°C

i = Vant hoff factor = 2 (For potassium bromide)

$K_b$ = molal boiling point elevation constant = 0.82°C/m

$m_{solute}$ = Given mass of solute (potassium bromide) = 54. g

$M_{solute}$ = Molar mass of solute (potassium bromide) = 119 g/mol

$W_{solvent}$ = Mass of solvent (liquid X) = 750. g

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.90=2\times 0.82^oC/m\times \frac{54\times 1000}{119g/mol\times 750}\\\\\text{Boiling point of solution}=119.9^oC=1.2\times 10^2^oC$

Hence, the boiling point of solution is $1.2\times 10^2^oC$

7 0

4 years ago

What is the density of the liquid in g/ml?

Mumz [18]

If you weigh 1.00 mL of liquid water it will have a mass of about 1 g so its density is about 1 g/mL or 1 g mL-1.

3 0

3 years ago