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vodomira [7]
3 years ago
10

a student performed an analysis of a sample for it's calcium content and got the following results: 14.92%, 14.91%, 14.88%, 14.9

1%. The actual amount of calcium in the sample is 15.70%. What conclusion can you draw about the accuracy and precision of these results?
Chemistry
1 answer:
kramer3 years ago
5 0
The question is asking to state the conclusion that can be formulated, base on the problem or the given analysis of the data, about the accuracy and precision  and base on my research, i would say that the Accuracy is Poor but the Precision is Good. I hope this would help you 
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Lead (II) nitrate reacts with ammonium carbonate to produce ammonium nitrate
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Pb(NO₃)₂ + (NH₄)₂CO₃ → PbCO₃ + 2 NH₄NO₃

Explanation:

Reaction of lead (II) nitrate with ammonium carbonate will produce lead (II) carbonate and ammonium nitrate.

The balanced chemical equation is:

Pb(NO₃)₂ + (NH₄)₂CO₃ → PbCO₃ + 2 NH₄NO₃

To balance the chemical equation the number of atoms of each element   entering the reaction have to be equal to the number of atoms of each   element leaving the reaction, in order to conserve the mass.

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3. Bacteria communicate with each other through electronic pulses.<br> O True<br> O False
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Why are there few fossils of protists. write a paragraph response
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Answer:

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1. (NH)2CrO-
Alex777 [14]

Question:

1. (NH)2CrO

a) Number of moles of H:

b) Number of moles of N:

Answer:

a) Number of moles of H: 2

b) Number of moles of N: 2

Explanation:

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Question:

2. Ag.SO.

a) Molar Mass:

b) Percent Composition of Ag:

c) Percent Composition of S:

d) Percent Composition of O:​

Answer:

a) Molar Mass:   155.93 Kg

b) Percent Composition of Ag:   69%

c) Percent Composition of S:  20.5%

d) Percent Composition of O:​ 10.2%

Explanation:

Molar mass =   molar mass of Ag +  molar mass of S +  molar mass of O  

=>107.87+32.06+16

=> 155.93 Kg

Percent Composition of Ag

= \frac{ \text{mass due  to Ag}}{\text {total molar mass}}  \times 100

= \frac{107.87}{155.93} \times 100

= 0.69 \times 100

= 69%

Percent Composition of S:

= \frac{ \text{mass due  to S}}{\text {total molar mass}}  \times 100

=\frac{32.06}{155.93} \times 100

= 0.205 \times 100

= 20.5%

Percent Composition of O:

= \frac{ \text{mass due  to O}}{\text {total molar mass}}  \times 100

= \frac{16.00}{155.93} \times 100

= 0.102 \times 100

= 10.2%

3 0
3 years ago
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