<u>Given:</u>
Mass of calcium nitrate (Ca(NO3)2) = 96.1 g
<u>To determine:</u>
Theoretical yield of calcium phosphate, Ca3(PO4)2
<u>Explanation:</u>
Balanced Chemical reaction-
3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2
Based on the reaction stoichiometry:
3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2
Now,
Given mass of Ca(NO3)2 = 96.1 g
Molar mass of Ca(NO3)2 = 164 g/mol
# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles
Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles
Molar mass of Ca3(PO4)2 = 310 g/mol
Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g
Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g
Molarity is defined as the number of moles of solute in 1 L of solution
mass fo KCl in the solution is - 5.0 g
number of moles of KCl - 5.0 g/ 74.5 g/mol = 0.067 mol
number of moles of KCl in 100 mL - 0.067 mol
therefore number of KCl moles in 1 L - 0.067 / 100 mL x 1000 mL = 0.67 M
molarity of KCl is 0.67 M
Lyman Series Working Formula:
1/λ = RH (1-(1/n^2))
Given:
n = 6
RH = Rydberg's constant = 1.0968x10^7 m^-1
c = speed of light = 3x10^8 m/s
Required:
Frequency (Hertz or cycles per second)
Solution:
To solve for the wavelength λ, we substitute the given in the working formula
1/λ = RH (1-(1/n^2))
1/λ = 1.0968x10^7 m^-1 (1-(1/6^2))
λ = 0.0000000938 m or 93.8 nm
To get the frequency, we will use the formula below.
f = c/λ
We then substitute c or the speed of light,
f = (3x10^8 m/s) / 0.0000000938 m
Therefore,
f = 3.2x10^15 s^-1
<em>ANSWER: Frequency = </em><em /><em>3.2x10^15 s^-1</em>
Answer:
GP.E = 5880 j
Explanation:
Given data:
Mass = 75 kg
height = 8 m
Potential energy = ?
Solution:
The formula for gravitational potential energy is
GPE = mgh
m = mass in kilogram
g = acceleration due to gravity
h = height in meter above the ground
Formula:
GP.E = mgh
Now we will put the values in formula.
g = 9.8 m/s²
GP.E = 75 Kg × 9.8 m/s²× 8 m
GP.E = 5880 Kg.m²/s²
Kg.m²/s² = j
GP.E = 5880 j
<span>Percentage
by mass is the amount in mass of a component in a mixture per 100 unit of mass of the
total mixture. Percentage by mass is the same as %w/w. We can determine this by dividing the mass of the solute with the total mass of the mixture. However, from the problem statement, we are given the volume of the water so there is a need to convert this value to mass by using the density of water. We calculate as follows:
Mass of solution = 100 mL (0.99993 g/mL) water + 25 g EtOH
Mass of solution = 124.993 g solution
%w/w = 25 g / 124.993 g x100
%w/w = 20% of EtOH</span>
