Answer:
See explanation below
Explanation:
First, you are not providing any data to solve this, so I'm gonna use some that I used a few days ago in the same question. Then, you can go and replace the data you have with the procedure here
The concentration of liquid sodium will be 8.5 MJ of energy, and I will assume that the temperature will not be increased more than 15 °C.
The expression to calculate the amount of energy is:
Q = m * cp * dT
Where: m: moles needed
cp: specific heat of the substance. The cp of liquid sodium reported is 30.8 J/ K mole
Replacing all the data in the above formula, and solving for m we have:
m = Q / cp * dT
dT is the increase of temperature. so 15 ° C is the same change for 15 K.
We also need to know that 1 MJ is 1x10^6 J,
so replacing all data:
m = 8.5 * 1x10^6 J / 30.8 J/K mole * 15 m = 18,398.27 moles
The molar mass of sodium is 22.95 g/mol so the mass is:
mass = 18,398.27 * 22.95 = 422,240.26 g or simply 422 kg rounded.
Answer:
It covers changes to the position of equilibrium if you change concentration, pressure or temperature. ... If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change
Explanation:
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
Answer:
Empirical formula is CH₄
Molecular formula = C₂H₈
Explanation:
Mass of carbon = 37.5 g
Mass of hydrogen = 12.5 g
Molecular weight = 32 g/mol
Molecular formula = ?
Empirical formula = ?
Solution:
Number of gram atoms of C = 37.5 g /12g/mol = 3.125
Number of gram atoms of H = 12.5 g / 1.008 g/mol= 12.4
Atomic ratio:
C : H
3.125/3.125 : 12.4 /3.125
1 : 4
C : H : = 1 : 4
Empirical formula is CH₄
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 32 / 16
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 ( CH₄)
Molecular formula = C₂H₈
