Answer:
1 mole of C2H6.
Explanation:
The balanced equation for the reaction is given below:
2C2H6 + 7O2 —> 4CO2 + 6H2O
We can determine the number of mole of C2H6 that reacted to produce 2 moles of CO2 as follow:
From the balanced equation above,
2 moles of C2H6 reacted to produce 4 moles of CO2.
Therefore, Xmol of C2H6 will react to produce 2 moles of CO2 i.e
Xmol of CO2 = (2 x 2)/4
Xmol of CO2 = 1 mole.
Therefore, 1 mole of C2H6 is required to produce 2 moles of CO2.
Molarity is defined as the number of moles of solute in 1 L of solution
the mass of Ca(NO₃)₂ present - 8.50 g
therefore number of moles of Ca(NO₃)₂ - 8.50 g / 164 g/mol = 0.0518 mol
the volume of solution prepared is 755 mL
therefore if there are 0.0518 mol in 755 mL
then in 1000 mL the number of moles - 0.0518 mol / 0.755 L
molarity is therefore - 0.0686 M
Answer:
8.44 atm
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2.25 L
Initial temperature (T₁) = 350 K
Initial pressure (P₁) = 1.75 atm
Final volume (V₂) = 1 L
Final temperature (T₂) = 750 K
Final pressure (P₂) =?
The final pressure of the gas can be obtained as illustrated below:
P₁V₁/T₁ = P₂V₂/T₂
1.75 × 2.25 / 350 = P₂ × 1 / 750
3.9375 / 350 = P₂ / 750
Cross multiply
350 × P₂ = 3.9375 × 750
350 × P₂ = 2953.125
Divide both side by 350
P₂ = 2953.125 / 350
P₂ = 8.44 atm
Thus, the final pressure of the gas is 8.44 atm.
Answer:

Explanation:
Hello there!
In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

Thus, by solving for the pressure of nitrogen and using consistent units, we obtain:

Answer: 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water. if 124 grams of the sodium acetate dissolves in 100 milliliters of water, then 6.2 grams of the sodium acetate can dissolve in 5 milliliters of water.