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Ghella [55]
3 years ago
10

Which of the following best represents both accuracy and precision

Chemistry
2 answers:
liq [111]3 years ago
6 0

Answer: Measurement B

Explanation:

Precision refers to the closeness of two or more measurements to each other.

For Example: If we measure a length three times and we get 3.5 m each time. Then the measurement is very precise.

Accuracy refers to the closeness of a measured value to a standard or known value.

For Example: If the length of the substance is 4 m and one person measures 3.9 m and another person measures 3.8 m. Then, the length measured by first person is more accurate.

Thus measurement B represents both accuracy as well as precision as the darts are close and pointed at the center.

makkiz [27]3 years ago
5 0
Accuracy is when you hit as close as to the target as you can and precision is when you are on point
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How many grams of precipitate will be formed when 20.5 mL of 0.800 M
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Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

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