Hello!
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
We have the following data:
m (mass) = ?
n (number of moles) = 5.20 moles
MM (Molar mass of C6H12) ≈ 84.2 g/mol
Now, let's find the mass, knowing that:
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when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
Answer:
C.
Explanation:
If the students want to know at what percent of CO2 in the air the plant will grow at the fastest, then the percent of CO2 should be a different value for each plant in the table.
There are 2 tables that have different values for the CO2 - the tables in answer choices C and D.
Since the students only want to know how the amount of CO2 affects the plant, every other variable should remain constant.
The only answer choice that has a changing value for the percent of CO2 and a constant value for every other variable is C.
