Answer:
53.1 mL
Explanation:
Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.
For the ideal gas law:
P1*V1/T1 = P2*V2/T2
Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.
At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:
P1 = 92.5 - 2.8 = 89.7 kPa
T1 = 23°C + 273 = 296 K
89.7*65/296 = 101.325*V2/273
101.325V2 = 5377.45
V2 = 53.1 mL
Answer:
50.76 mol H2O.
Explanation:
The photosynthesis follows the equation:
6CO2 + 6H2O ---> C6H12O6 + 6O2
This means that 6 mol of H2O are needed to obtain 1 mol of C6H12O6 (see the numbers that precedes every molecule to know how many mols are in game).
So we can say that:
1 mol C6H12O6 --------- 6 mol H2O
8.46 mol C6H12O6 -----x= 8.46 x 6 : 1 = 50.76 mol H20
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
Explanation:
solar and energy i had to answer this one before
To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the <span>∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.</span>
