Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter
Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.
The algebraic expression would be:
<em>ppb [=] micrograms of compound/liter of solution</em>
We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.
For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.
The respective procedure is in a attached file.
M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769
Answer:
Explanation:
For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..
Answer:
The atomic mass of phosphorus is 29.864 amu.
Explanation:
Given data:
Atomic mass of phosphorus = ?
Percent abundance of P-29 = 35.5%
percent abundance of P-30 = 42.6%
Percent abundance of P-31 = 21.9%
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100
Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100
Average atomic mass = 1029.5 + 1278 + 678.9/ 100
Average atomic mass = 2986.4 / 100
Average atomic mass = 29.864 amu.
The atomic mass of phosphorus is 29.864 amu.
