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notsponge [240]
3 years ago
12

You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions

via serial dilution. 1. Perform calculations to determine the volume of the 1.0x10-5 M stock solution needed to prepare 10.0 mL of a 2.0x10-6 M solution. Perform calculations to determine the volume of the 2.0x10-6 M solution needed to prepare 10.0 mL of a 5.0x10-7 M solution. You are welcome to do the calculations in your notebook prior to lab. 2. Prepare 10.0 mL of a 2.0x10-6 M solution using a graduated cylinder. Transfer the solution to a clean 50 mL beaker. 3. Prepare 10.0 mL of a 5.0x10-7 M solution using a graduated cylinder. Transfer the solution to a clean 50 mL beaker.
Chemistry
1 answer:
artcher [175]3 years ago
8 0

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

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Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

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4 0
3 years ago
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Answer:

We need 247 mL of NaOH

Explanation:

Step 1: Data given

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