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4 years ago

Which of the following is an example of a scientific law?

Chemistry

2 answers:

Tresset [83]4 years ago

8 0

Answer:

Experimental results

Bogdan [553]4 years ago

7 0

Experimental results that show that d=m/v

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7) Calculate the number of moles of Chlorine gas if 50.0L of the gas is at 0.50atm of pressure at 400K.

Ostrovityanka [42]

Answer:

so i never did this befor so i am sorry hope you get the answer

Explanation:

7 0

3 years ago

A golfer hits a golf ball with a club. The mass of the ball is 0.05 kg. The ball accelerates at

Alex73 [517]

Answer:

100 n im just guess

Explanation:

if it helps.

have a good day

hope you stay safe

4 0

3 years ago

If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised

alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}$

Solving for V₂ , we get:

V₂ = 15.04 mL

3 0

3 years ago

A gas initial volume and pressure is 5m^3 and 101320 Pa and allows to expand up to 12m^3 then what is the final pressure?

Anton [14]

Answer:

New pressure = 42216.66 Pa

Explanation:

Given that,

Initial volume, V₁ = 5 m³

Final pressure, P₁ = 101320 Pa

Final volume, V₂ = 12 m³

We need to find the final pressure of the gas. We know that the relation between pressure and volume is given by :

$P\propto \dfrac{1}{V}\\\\P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{101320 \times 5}{12}\\\\P_2=42216.67\ Pa$

So, the new pressure is equal to 42216.66 Pa.

6 0

3 years ago

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$