Molar mass of 13c = 13 grams
number of moles = mass / molar mass
therefore,
number of moles = 7 / 13
To know the number of atoms in 7/13 moles, we simply multiply the number of moles by Avogadro's number as follows:
number of atoms = (7/13) x 6.022 x 10^23 = 3.2426 x 10^23 atoms
Answer:
Mass = 90.28 g
Explanation:
Given data:
Mass of Ca(OH)₂ = ?
Volume of solution= 1.5 L
Molarity of solution = 0.81 M
Solution:
First of all we will calculate number of moles.
Molarity = number of moles / volume in L
by putting values,
0.81 M = Number of moles / 1.5 L
Number of moles = 0.81 M × 1.5 L
Number of moles = 1.22 mol
Mass of Ca(OH)₂ in gram:
Mass = number of moles × molar mass
Mass = 1.22 mol × 74.09 g/mol
Mass = 90.28 g
Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature
PV = k
where P - pressure , V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
1.25 atm x 0.75 L = P x 1.1 L
P = 0.85 atm
final pressure is B) 0.85 atm
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
