A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.
<h3>What is "ppm"?</h3>
"ppm" of "parts per million" is a unit of concentration equivalent to milligrams of solute per liters of solution.
A pool must maintain a chlorine concentration of 3.0 ppm (3.0 mg/L). The mass of chlorine in 3.4 × 10⁶ L is:
3.0 mg Cl₂/L × 3.4 × 10⁶ L = 1.0 × 10⁷ mg Cl₂
A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.
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Answer:
Explanation:
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In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:
Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:
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Molarity = moles of solute(HCl)
------------------------------------
volume of the solution
= 1
------
5
= 0.2M.
Hence option B is correct.
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Answer:
67.5%
Explanation:
Step 1: Write the balanced equation for the electrolysis of water
2 H₂O ⇒ 2 H₂ + O₂
Step 2: Calculate the theoretical yield of O₂ from 17.0 g of H₂O
According to the balanced equation, the mass ratio of H₂O to O₂ is 36.04:32.00.
17.0 g H₂O × 32.00 g O₂/36.04 g H₂O = 15.1 g O₂
Step 3: Calculate the percent yield of O₂
Given the experimental yield of O₂ is 10.2 g, we can calculate its percent yield using the following expression.
%yield = (exp yield / theoret yield) × 100%
%yield = (10.2 g / 15.1 g) × 100% = 67.5%
