Molar mass CO2 = 12 + 16 x 2 => 44.0 g/mol
1 mol ----------- 44.0 g
2.1 mol --------- ( mass CO2)
mass CO2 = 2.1 x 44.0 / 1
mass CO2 = 92.4 / 1
= 92.4 g of CO2
hope this helps!
HC₂H₃O₂ , equilibrium can be represented as:
HC₂H₃O₂ + H₂O--------->CH₃COO⁻ + H₃O⁺
HC₂H₃O₂, that is acetic acid dissociate in water that is H₂O and formed acetate anion that is CH₃COO⁻ and hydronium ions that is H₃O⁺.
Its equiibrium constant can be represented as:
Ka=\frac{[CH_{3}COO^{-1}]\times [H_{3}O^{+1}]}{CH_{3}COOH}
While NaC₂H₃O₂ can be dissociate as:
NaC₂H₃O₂------>CH₃COO⁻ + Na⁺
Sodium acetate that is NaC₂H₃O₂, dissociates as forming acetate anion and sodium cation.
Answer:
The average atomic mass of bromine is 79.9 amu.
Explanation:
Given data:
Percentage of Br⁷⁹ = 55%
Percentage of Br⁸¹ = 45%
Average atomic mass of bromine = ?
Formula:
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
Now we will put the values in formula.
Average atomic mass = [55 × 79] + [81 ×45] / 100
Average atomic mass = 4345 + 3645 / 100
Average atomic mass = 7990 / 100
Average atomic mass = 79.9 amu
The average atomic mass of bromine is 79.9 amu.
1) You neeed to know and use the Ksp for BaF2.
At 25°C this Ksp is 1.0 * 10 ^ - 6
2) The solutibility of BaF2 is given by:
BaF2 ⇄ Ba(2+) + 2F(-)
x 2x
=> Ksp = x * (2x)^2 = 4x^3
3) When you have a NaF solution, you have to take into accout the concentration of the NaF solution
M = 0.1
Now the equilibrium species are:
BaF2 ⇄ Ba(2+) + 2F(-)
x 2x + 0.10
And Ksp = x* [2x + 0.10]^2 = 1.0 * 10 ^ -6
Given that the Ksp << 1 you may assume that 2x << 0.1 => 2x + 0.1 ≈ 0.1
=> 1.0 * 10 ^ - 6 ≈ x(0.1)^2 = 0.01x
=> x = 1.0 * 10^ -6 / 0.01 = 1.0 * 10^ - 4 M = 0.0001 M
That is the molar solubility.
4) Now, you calculate the number of moles from the molarity's formula:
M = n / v => n = M * v = 0.0001 M * 0.500 l = 0.00005 mol
And now convert to grams,
mass in grams = number of moles * molar mass
molar mass of BaF2 = 175.34 g/mol
mass in grams = 0.00005 moles * 175.34 g / mol = 0.0088 g
Answer: 0.0088 g
Its The Ribosomes That Define That Type Of Cell