Answer:
(a) 17,178 mg/m3
(b) 11,625 mg/m3
Explanation:
The concentration of CO in mg/m3 can be calculated as
For standard conditions (1 atm and 25°C), P/RT is 0.0409.
Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.
The molecular weigth of CO is 28 g/mol.
(1) For 25°C and 1 atm conditions
(b) For 200°C and 1.1 atm,
Then the concentration in mg/m3 is
Answer:
[NaCH₃COO] = 2.26M
Explanation:
17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)
Let's determine the volume of solution, by density
Mass of solution / Volume of solution = Solution density
100 g / Volume of solution = 1.09 g/mL
100 g / 1.09 g/mL = 91.7 mL
17 grams of solute is contained in 91.7 mL
Molarity (M) = Mol of solute /L of solution
91.7 mL / 1000 = 0.0917L
17 g / 82 g/m = 0.207 moles
Molariy = 0.207 moles / 0.0917L → 2.26M
Answer:
The elements in the reactants are the same as the elements in the products.
Explanation:
You can't have more or less elements or atoms, but you also can't just have the same number of atoms because then it could be different elements, which can't happen.
Answer:
6.31g/mol
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
Mole (n) = mass (m)/molar mass (Mm)
* Mm = m/n
Also, density (p) = mass (m) ÷ volume (V)
PV = nRT
Since n = M/Mm
PV = M/Mm. RT
PV × Mm = m × RT
Divide both sides by V
P × Mm = m/V × RT
Since p = m/V
P × Mm = p × RT
Mm = p × RT/P
Mm = 0.249 × 0.0821 × 293/0.95
Mm = 5.989 ÷ 0.95
Mm = 6.31g/mol
