The answer is: 231.25 ppm.
To solve this, compute first the percentage of hydrogen in the 3.2 g air sample. % = (0.00074g/3.2g)*100 = 0.023125%
1% = 10,000ppm <--- use this as conversion factor.
0.023125%(10,000ppm/1%) = 231.25 ppm
The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.
<h3>How to calculate the effluent flow?</h3>
It should be noted that the total inflow will be equal to the total outflow. Therefore,
0.2 + 0.048 = 0.05 + We
Collect like terms
Qe = 0.2 + 0.048 - 0.05
Qe = 0.198m³/sec
The concentration will be:
= (360 × 1000)/0.05
= 7200mg/L.
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Answer
is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.
<span>
Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] =
3s(Ca₃(PO₄)₂) =
3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.
s = ⁵√(Ksp ÷ 108).
The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ =
→ V₂ = 
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL
Answer: Argon (Ar), which has 18 protons.