If the observed value for a density is 0.80 g/mL and the accepted value is 0.70 g/mL what is the percent error?

Chemistry

1 answer:

kvv77 [185]3 years ago

4 0

Answer:

<h2>The answer is 14.29 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual density = 0.70 g/mL

error = 0.8 - 0.7 = 0.1

So we have

$P(\%) = \frac{0.1}{0.7} \times 100 \\ = 14.285714...$

We have the final answer as

Hope this helps you

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Consider the reaction cacn2 + 3 h2o → caco3 + 2 nh3 . how much caco3 is produced if 47.5 moles nh3 are produced? 1. 0.950 g 2. 9

zhannawk [14.2K]

The CaCO3 produced if 47.5 moles of NH3 produced is calculated as follows

CaCN2 +3H2O = CaCO3 + 2NH3

by use of mole ratio between CaCO3 to NH3 which is 1:2 the moles of CaCO3 is therefore = 47.5 /2= 23.75 moles

mass of CaCO3 is therefore = moles x molar mass
= 23.75 moles x 100g/mol= 2375 grams which is approximate 2380 grams(answer 6)

8 0

4 years ago

Indicate what happens to the concentration of Pb2+ in each half cell

Taya2010 [7]

Answer:

Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below

The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.

Explanation:

The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.

In the Left Beaker (Left half cell), their is less concentration

Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side

In the Right Beaker (right half cell), their is more concentration

Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side

Electrons will flow from Left to Right direction.