Answer:
Volume required = 0.327 L
Explanation:
Given data:
Volume in L = ?
Molarity of solution = 1.772 M
Mass of BaCl₂ = 123 g
Solution:
First of all we will calculate the number of moles of BaCl₂,
Number of moles = mass/molar mass
Number of moles = 123 g/ 208.23 g/mol
Number of moles = 0.58 mol
Now, given problem will solve by using molarity formula.
Molarity = number of moles / volume in L
1.772 M = 0.58 mol / Volume in L
Volume in L = 0.58 mol / 1.772 M
Volume in L = 0.327 L
Answer:
1.23 j/g. °C
Explanation:
Given data:
Mass of metal = 35.0 g
Initial temperature = 21 °C
Final temperature = 52°C
Amount of heat absorbed = 320 cal (320 ×4.184 = 1338.88 j)
Specific heat capacity of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 52°C - 21 °C
ΔT = 31°C
1338.88 j= 35 g ×c× 31°C
1338.88 j= 1085 g.°C ×c
1338.88 j/1085 g.°C = c
1.23 j/g. °C = c
3.25 kg in g = 3.25 * 1000 = 3250 g
Molar mass C₂H₆O₂ = 62.0 g/mol
Mass solvent = 7.75 kg
Number of moles:
n = mass solute / molar mass
n = 3250 / 62.0
n = 52.419 moles
Molality = moles of solute / kilograms of solvent
M = 52.419 / 7.75
M = 6.7637 mol/kg
hope this helps!
The pH scale is a measurement system that indicates the concentration of H+ ions in a particular solution.
We are given the amount of Nitrogen gas and hydrogen gas reacted to form ammonia:
N2 = 19.25 grams
H2 = 11.35 grams
Set-up a balanced chemical equation:
N2 + 3H2 ==> 2NH3
The theoretical amount of ammonia that will be produced from the given amounts is:
First, we need to determine the limiting reactant to serve as our basis for calculation.
number of moles / stoichiometric ratio
N2 = 19.25 g/ 28 g/mol / 1 = 0.6875
H2 = 11.35 g/ 2 g/mol /3 = 1.89
The limiting reactant is N2.
0.6875 moles N2 * (2 NH3/ 1 N2) * 17 g/mol NH3
The amount of NH3 produced is 23.375 grams of ammonia. <span />
