Answer:
138 mg
Explanation:
A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.
Step 1: Convert the mass of drinking water to kilograms.
We will use the relation 1 kg = 1000 g.
Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water
Hey it isnt letting me submit my answer to your question on the Japanese chart, so imma just submit it here
<span>The answer is 8. Hope this help!!!
</span>
A. M x L = moles.
<span>b. CH3COOH + NaOH ==> CH3COONa + H2O </span>
<span>I...6 mmols....0.......7.5 mmoles </span>
<span>C... 0........0.51 mmols..0 </span>
<span>E...6-0.511 ....0.......7.5+0.511 </span>
<span>I stands for initial </span>
<span>C stands for change. </span>
<span>E stands for equilibrium. </span>
<span>Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros. </span>
<span>c. done as in b.</span>
Answer:
0.171 M
Explanation:
Step 1: Given data
- Mass of H₃PO₄ (solute): 3.35 g
- Volume of solution (V): 200 mL
Step 2: Calculate the moles of solute
The molar mass of H₃PO₄ is 97.99 g/mol.
3.35 g × 1 mol/97.99 g = 0.0342 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
200 mL × 1 L/1000 mL = 0.200 L
Step 4: Calculate the molarity of the solution
We will use the definition of molarity.
M = moles of solute / liters of solution
M = 0.0342 mol/0.200 L = 0.171 M
