The sugar reacts with the gas, turning it to a semi-solid and sticky substance; clogging the gas lines along with many other things.
Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:
Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration
Molar mass:
HF = 1 + 19 = 20.0 g/mol
Number of moles :
124 / 20.0 => 6.2 moles
Volume = 2.4 L
M = n / V
M = 6.2 / 2.4
M = 2.6 M
Answer A
hope this helps!
Answer:
Principle quantum number.
Explanation:
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,
cC ⇔ aA + bB
Ksp = {[A^a][B^b]}
Now, for the this problem, the reaction is
BaSO₄ ⇔ Ba²⁺ + SO₄²⁻
The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.
1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
