Mass of H2C2O4 :
mm = 90.04 g/mol
number of moles : 0.0223 moles
m = n * mm
m = 0.0223 * 90.04
m = 2.007 g
hope this helps!.
Answer:
The atomic mass of gallium (Ga) = <u>69.723 g/mol</u>
Explanation:
Given: Two isotopes of Gallium (Ga) are Gallium-69 (⁶⁹Ga) and Gallium-71 (⁷¹Ga)
<u>For ⁶⁹Ga: </u>
Relative abundance = 60.12% = 60.12 ÷ 100 = 0.6012; Atomic mass = 68.9257 g/mol
<u>For ⁷¹Ga:</u>
Relative abundance = 39.88% = 39.88 ÷ 100 = 0.3988; Atomic mass = 70.9249 g/mol
∴ The atomic mass of Ga = (Relative abundance of ⁶⁹Ga × Atomic mass of ⁶⁹Ga) + (Relative abundance of ⁷¹Ga × Atomic mass of ⁷¹Ga)
⇒ Atomic mass of Ga = (0.6012 × 68.9257 g/mol) + (0.3988 × 70.9249 g/mol) = <u>69.723 g/mol</u>
<u>Therefore, the atomic mass of gallium (Ga) = 69.723 g/mol</u>
This would be c as for the amswer
The question is missing the data sets.
This is the complete question:
A single penny has a mass of 2.5 g. Abbie and James
each measure the mass of a penny multiple times. Which statement about
these data sets is true?
O Abbie's measurements are both more accurate
and more precise than James'.
O Abbie's measurements are more accurate,
but less precise, than James'.
O Abbie's measurements are more precise,
but less accurate, than James'.
O Abbie’s measurements are both less
accurate and less precise than James'.
Penny masses (g)
Abbie’s data
2.5, 2.4, 2.3, 2.4, 2.5, 2.6, 2.6
James’ data
2.4, 3.0, 3.3, 2.2, 2.9, 3.8, 2.9
Answer: first option, Abbie's measurements are both more accurate
and more precise than James'.
Explanation:
1) To answer this question, you first must understand the difference between precision and accuracy.
<span>Accuracy is how close the data are to the true or accepted value.
</span>
<span>Precision is how close are the data among them, this is the reproducibility of the values.</span>
Then, you can measure the accuracy by comparing the means (averages) with the actual mass of a penny 2.5 g.
And you measure the precision by comparing a measure of spread, as it can be the standard deviation.
2) These are the calculations:
Abbie’s data
Average: ∑ of the values / number of values
Average = [2.5 + 2.4 + 2.3 + 2.4 + 2.5 + 2.6 + 2.6 ] / 7 = 2.47 ≈ 2.5
Standard deviation: √ [ ∑ (x - mean)² / (n - 1) ] = 0.11
James’ data
Average = [2.4 + 3.0 + 3.3 + 2.2 + 2.9 + 3.8 + 2.9] / 7 = 2.56 ≈ 2.6
Standard deviation = 0.53
3) Conclusions:
1) The average of Abbie's data are closer to the accepted value 2.5g, so they are more accurate.
2) The standard deviation of Abbie's data is smaller than that of Jame's data, so the Abbie's data are more precise.