Answer:
Explanation: n=m/M(molar mass)
n=24.3 grams/(16+2x1.008)grams/moles(molar mass of H2O)
n=24.3grams/18.016grams/moles
n=1.35moles
Answer:
AgCl + H2 - Chemical Equation Balancer.
Answer:
A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :
<u>2.67%</u>
Explanation:
Note : Look at the density of potassium nitrate in water if given in the question.
<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>
Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>
Density of potassium nitrate in water KNO3 = 2.11 g/mL

Density = 2.11 g/mL
Volume of solution = 86.4 mL



Mass of Solute = 4.87 g
Mass of Solution = 183.2 g
w/w% of the solute =


w/w%=2.67%
Answer:
-973 KJ
Explanation:
The balanced reaction equation is;
N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)
Reduction potential of hydrazine = -1.16 V
Reduction potential of chlorine = 1.36 V
From;
E°cell= E°cathode - E°anode
E°cell= 1.36 - (-1.16)
E°cell= 2.52 V
∆G°=- nFE°cell
n= number of moles of electrons = 4
F= Faraday's constant = 96500 C
E°cell = 2.52 V
∆G°=- (4 × 96500 × 2.52)
∆G°= -972720 J
∆G°= -972.72 KJ