Answer:
The enthalpy of the solution is -35.9 kJ/mol
Explanation:
<u>Step 1:</u> Data given
Mass of lithiumchloride = 3.00 grams
Volume of water = 100 mL
Change in temperature = 6.09 °C
<u>Step 2:</u> Calculate mass of water
Mass of water = 1g/mL * 100 mL = 100 grams
<u>Step 3:</u> Calculate heat
q = m*c*ΔT
with m = the mass of water = 100 grams
with c = the heat capacity = 4.184 J/g°C
with ΔT = the chgange in temperature = 6.09 °C
q = 100 grams * 4.184 J/g°C * 6.09 °C
q =2548.1 J
<u>Step 4:</u> Calculate moles lithiumchloride
Moles LiCl = mass LiCl / Molar mass LiCl
Moles LiCl = 3 grams / 42.394 g/mol
Moles LiCl = 0.071 moles
<u>Step 5:</u> Calculate enthalpy of solution
ΔH = 2548.1 J /0.071 moles
ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)
The enthalpy of the solution is -35.9 kJ/mol
Answer:
See explanation
Explanation:
Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.
In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.
Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.
Answer:
The coefficient for PH3 is 8. Option D is correct.
Explanation:
Step 1: The unbalanced equation
P2H4(g) ⇆ PH3(g) + P4(s)
Step 2: Balancing the equation
P2H4(g) ⇆ PH3(g) + P4(s)
On the left side we have 4x H (in P2H4), on the right side we have 3x H (in PH3). To balance the amount of H on both sides, we have to multiply P2H4 on the left side by 3 and PH3 on the right by 4.
3P2H4(g
) ⇆ 4PH3(g) + P4(s)
On the left side we have 6x P (in 3P2H4) on the right side we have 8x P (4x in 4PH3 and 4x in P4). To balance the amount of P on bot hsides, we have to multiply 3P2H4 by 2 and 4PH3 also by 2. Now the equation is balanced
6P2H4(g
) ⇆ 8PH3(g) + P4(s)
The coefficient for PH3 is 8. Option D is correct.
Answer:
The percent by mass of nitric acid in the mixture is 5.48 %
Explanation:
Givn that
Mass of HNO3 = 9.03 grams
Volume of KOH = 12.0 mL = 0. 012 L
Molarity of KOH = 0.655 M
The balanced equation
Ba(OH)2 + HNO3 → Ba(NO3)2 + H2O
Calculate the moles of KOH
Moles of Ba(OH)2 = molarity KOH * volume
Moles Ba(OH)2 = 0.655 M * 0.012 L
Moles Ba(OH)2 = 0.00786 moles
Calculate moles of HNO3
For 1 mol of Ba(OH)2 we need 1 mol of HNO3
For 0.00786 moles of Ba(OH)2 we need 0.00786 moles of HNO3
Calculate mass of HNO3
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.00786 moles * 63.01 g/mol
Mass HNO3 = 0.495 grams
Calculate mass % HNO3 in sample
mass % = (0.495 grams / 9.03 grams)*100%
mass % = 5.48 %
The percent by mass of nitric acid in the mixture is 5.48 %