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prohojiy [21]
3 years ago
15

If the initial concentration of a is 0.00540 m, what will be the concentration after 795 s?

Chemistry
1 answer:
zavuch27 [327]3 years ago
7 0
<span>Feb 19, 2014 - The units of k tell you that this is a second order reaction. So, to solve this, you need to use the integrated rate law for a 2nd order reaction: 1/[A] = kt + 1/[A]o 1/[A] = 0.540/Ms (835 s) + 1/0.00640 1/[A] = 607 [A] = 1.65X10^-3 M.</span><span>
</span>
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Phosphates are formed when phosphorus reacts with
777dan777 [17]

Answer: nitrogen

Explanation: bc

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Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti
djverab [1.8K]
<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

                        = 80,000 g

Therefore;

Moles of water = \frac{80,000g}{18.02 g/mol}

                = 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O  available = 65.0 kg or 65,000 g

Molar mass Li₂O  = 29.88 g/mol

Moles of Li₂O  = 65,000 g ÷ 29.88 g/mol

          = 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

  • Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
  • However, the number of moles of water to be removed is 4.44 × 10³ moles  but only 2.175 × 10³ moles will react with the available Li₂O.
  • This means, Li₂O  is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water =  4.44 × 10³ moles  - 2.175 × 10³ moles

                                           = 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

                                          = 4.0815 × 10⁴ g or

                                          = 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0
3 years ago
What is the pH of a solution with a 3.8 × 10−4 M hydronium ion concentration?
mamaluj [8]

Answer:

3.4

Explanation:

The pH scale is used to express the acidity or basicity of a solution.

  • If the pH < 7, the solution is acid.
  • If the pH = 7, the solution is neutral.
  • If the pH > 7, the solution is basic.

Given the hydronium ion concentration [H₃O⁺] = 3.8 × 10⁻⁴ M, we can calculate the pH using the following expression.

pH = -log [H₃O⁺]

pH = -log 3.8 × 10⁻⁴

pH = 3.4

This solution is acid.

8 0
2 years ago
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0.600
D) 0.600 is the final concentration of the solution of KCl.
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6 0
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Explanation:

2H2+O2------->2H2O

its yr balanced equation.

hope it helps

<h2>stay safe healthy and happy....</h2>
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