Answer:
[Ba(OH)₂] = 0.21 M
Explanation:
Molarity is a sort of concentration for solute that determines the moles of solute contained in 1 L of solution (mol/L).
Solution's volume = 45.6 mL
We convert the volume from mL to L → 45.6 mL . 1 L/1000 mL = 0.0456L
Our solute: Ba(OH)₂
We convert the mass of solute to moles → 1.68 g / 171.33 g/mol =
0.00980 moles
Molarity → mol/L → 0.0980 moles / 0.0456L = 0.21 M
I am pretty sure it is nitrogen, oxygen, carbon and hydrogen.
Answer:
Protons: 84
Neutrons: 127
Explanation:
The number of protons is the atomic number. The number of neutrons can be found by subtracting the proton number (84) from the isotope number (211)
<span>83.9%
First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements.
Atomic weight aluminum = 26.981539
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol
Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol
Now determine how many moles of C6H6 was produced
Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol
Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required.
0.012174851 mol / 3 = 0.004058284 mol
Then multiply by the molar mass to get the number of grams that was originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So
1.048227218 g / 1.25 g = 0.838581775 = 83.8581775%
And of course, round to 3 significant digits, giving 83.9%</span>
Answer:
68.6 °C
Explanation:
From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'
Q = Q'
Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g
Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown
So, mL = m'c(θ₂ - θ₁)
θ₂ - θ₁ = mL/m'c
θ₂ = mL/m'c + θ₁
substituting the values of the variables into the equation, we have
θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C
θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C
θ₂ = 43.56 °C + 25 °C
θ₂ = 68.56 °C
θ₂ ≅ 68.6 °C
So, the final temperature (in °C) of the metal block is 68.6 °C.