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nexus9112 [7]
3 years ago
10

It is dangerous to have over 1.3 ppm of copper (Cu) in drinking water. A small town investigates their water supply and finds th

ey have 0.002 g of copper for every 1000 g of solution. Should the town find a new source of water
Chemistry
1 answer:
lianna [129]3 years ago
3 0

Answer:

The town should find a new source of water.

Explanation:

Step 1: Convert the mass of copper to milligrams

We will use the conversion factor 1 g = 1000 mg.

0.002 g × (1000 mg/1 g) = 2 mg

Step 2: Convert the mass of solution to kilograms

We will use the conversion factor 1 kg = 1000 g.

1000 g × (1 kg/1000 g) = 1 kg

Step 3: Calculate the concentration of Cu in ppm (mg/kg)

2 mg/1 kg = 2 ppm

This is over 1.3 ppm, so the town should find a new source of water.

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Calculate the mass of water producd from the reaction of 126g of pentaboron noahydride with 192g of moleculAR OXYGEN
Sindrei [870]

Answer:

81.08g of H_{2}O will be produced.

Explanation:

Write down the balanced chemical equation:

B_5H_9 + O_2 ⇒ B_2O_3+H_2O

2B_5H_9+12O_2 ⇒ 5B_2O_3+9H_2O

Determine the limiting reagent:

B_5H_9 :-    126/63.12646 = 1.995993 mol

               1.995993/2 = 0.9979965

O_2 :-         192/31.9988 = 6.000225 mol

               6.000225/12 = 0.50001875

Therefore, O_2, is the limiting reagent.

Use stoichiometry ratios to determine the number of moles of water produced:

         O_2         :        H_2O

         12          :          9

  6.000225   :       4,500168756328362

Use the mole formula to calculate the mass of water produced:

n=\frac{m}{M} \\m=nM\\m=(4.500168756328362)(18.01528)\\m=81.08g

5 0
3 years ago
Result of air movement speed up evaporation​
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4 0
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How many grams of NaOH would be required to make 1.0 L of a 1.5 M solution
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60 grams are required.

Hope this helped you!
4 0
3 years ago
Please help!!! Stoich Question
Greeley [361]

Answer:

4KO₂ + 2CO₂ -> 2K₂CO₃ + 3O₂

<u> Step 1: Find the moles of O₂.</u>

n(O₂) = mass/ Mr.

n(O₂) = 100 / 32 = 3.125 mol

<u>Step 2: Find the ratio between KO₂ and O₂.</u>

        <u>KO₂ </u>    :      <u> O₂</u>

          4        :       3

        4/3       :       1

  (4*3125)/3  :      3.125

=4.167 mol of KO₂

Thus now we know, to produce 100 g of O₂, we need 4.167mol of KO₂

<u>Step 3: Find the mass of KO₂:</u>

<u />

mass = mol * Mr. (KO₂)

Mass = 4.167* 71.1

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Are usually solid shiny and conductive object
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