What aspect of a maple tree might be studied in chemistry

Which gas has approximately the same density as c2h6 at stp no nh3 h20 so2?

zmey [24]

1 mole of any gas occupy 22.4 L at STP (standard temperature and pressure, 0°C and 1 atm).

Let given gases be 1 mole. So their volumes will be the same, 22.4 liters.

Density is the ratio of mass to volume.

By formula; density= mass/volume; d=m/V

To find out masses of gases, do the mole calculation.

By formula; mole= mass/molar mass; n= m/M; m= n*M

Molar masses are calculated as
1. C₂H₆ (ethane) = 2*12 g/mol + 6*1 g/mol= 30 g/mol
2. NO (nitrogen monoxide) = 1*14 g/mol + 1*16 g/mol= 30 g/mol
3. NH₃ (ammonia) = 1*14 g/mol + 3*1 g/mol= 17 g/mol
4. H₂O (water) = 2*1 g/mol + 1*16 g/mol= 18 g/mol
5. SO₂ (sulfur dioxide) = 1*32 g/mol + 2*16 g/mol= 64 g/mol
Use Periodic Table to get atomic mass of elements.

Since their volumes are equal, compounds having the same molar mass will have the same density.
Recall the formula d= m/V.

Ethane and nitrogen monoxide have the same density.

The answer is C₂H₆ and NO.

6 0

3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

Answer: The experimental van't Hoff factor is 1.21

Explanation:

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

Hence, the experimental van't Hoff factor is 1.21

7 0

3 years ago

PLEASE HELPPPPP I WILL GIVE BRAINLEST FOR WHOEVER FINDS THE ANSWER!!!

Elza [17]

I think that it is shear adhesion

5 0

3 years ago

Read 2 more answers

If you pump air into cycle tyre a slight warming effect is noticed at valve stem why

Mashutka [201]

As the air molecules move through the valve they have friction as they hit the walls, and its this friction that causes it to heat up.

6 0

4 years ago