<span>In 3.3 moles of potassium sulphide (K2S), there are 363.99 g. Let's first calculate the molar mass of K2S (Mr) which is the sum of atomic masses (Ar) of its elements. According to the periodic table, Ar(K) = 39.1 g/mol and Ar(S) = 32.1 g/mol. Mr(K2S) = 2Ar(K) + Ar(S) = 2 * 39.1 + 32.1 = 110.3 g/mol. Thus, there are 110.3 g per 1 mol. There will be x grams in 3.3 moles. 110.3 g : 1 mol = x : 3.3 mol. x = 110.3 g * 3.3 mol : 1 mol = 363.99 g.</span>
This answer is 24 because 2.17 x 10 -8 is 24 so that would be your answer
Answer : The reaction 2 is spontaneous.
Explanation :
As we know that:
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.
Reaction 1:
Glucose + Pi
glucose-6-phosphate + H₂O, ΔG = +13.8 kJ/mol
Reaction 2:
ATP + H₂O
ADP + Pi, ΔG = -30.5 kJ/mol
From this we conclude that the value of ΔG is negative. So, reaction 2 is a spontaneous reaction.
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)