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STatiana [176]
3 years ago
11

Following the procedure, a student recorded the initial volume of NaOH in his buret as 2.82 mL. He then titrated a 15.00 mL samp

le of 0.1027 M HCl until a persistent pale pink appeared. He recorded the final volume of NaOH in his buret as 17 mL. What is the concentration (M) of his NaOH solution?
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0

Answer:

The concentration of this NaOH solution was 0.104 M

Explanation:

<u>Step 1:</u> Data given

Initial volume of NaOH = 2.82 mL = 0.00282 L

The sample is titrated with 15.00 mL of 0.1027 M HCl

Final volume NaOH = 17 mL = 0.017 L

<u>Step 2:</u>  Calculate change in volumes

17 mL - 2.82 mL = 14.18 mL = 0.01418 L

<u>Step 3: </u>Calculate the concentration of the NaOH

C1*V1 = C2*V2

⇒ with C1 = the initial concentration = TO BE DETERMINED

⇒ with V1 = the initial volume =  0.0148 L

⇒ with C2 = the final concentration = 0.1027

⇒ with V2 = the final volume = 0.015 L

C1 = (0.1027*0.015) /0.0148

C1 = 0.104 M

The concentration of this NaOH solution was 0.104 M

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The amount of matter in a given amount of volume is referred to as
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Answer:

It would be Density because:

Density= mass/ volume

Density refers to the amount of matter or mass of a substance over a volume

7 0
3 years ago
How to write a balanced chemical equation from empirical formula?
netineya [11]
Let's use the example: H2O --->  H2 + O2
We find how many elements of a product are on one side and how many elements on the other side.
Reactant: H=2 O=1
Product:   H=2 O=2
We need to make the same amount of hydrogen and oxegyn atoms on each side, regardless of how high the numbers are, and we do this by adding coefficients to the compounds.

Reactant: H=4 O=2
Product  : H=4 O=2
2 H2O--->   2 H2 + O2
8 0
3 years ago
A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
Ulleksa [173]

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




7 0
3 years ago
Convert 23.064 mass to mg
Ierofanga [76]

ANSWER: 23064 mg

EXPLANATION: To convert grams to milligrams, we multiply by 1000.

23.064 g x 1000 = 23064 mg

3 0
3 years ago
What is the percent composition by mass for hydrogen in water (H2O)?
saw5 [17]

Explanation:

Percentage composition = 2/18 = 11.11%.

5 0
3 years ago
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