Answer:
a. 86,942 = 8.69x10⁴
f. 56,018 = 5.60x10⁴
b. 0.00621875 = 6.22x10⁻³
g. 0.003870 = 3.87x10⁻³
c. 107062.03 = 1.07x10⁵
h. 15.9994 = 16.0
d. 4.00049 = 4.00
i. 12.011 = 12.0
e. 96 = 9.60x10¹
j. 2,000 = 2.00x10³
Explanation:
Hello,
In this case, rounding off each measurement to three significant figures, considering that if the fourth figure is five or more, the third one is rounded, we obtain:
a. 86,942 = 8.69x10⁴ (four places until the last digit and third digit is not rounded).
f. 56,018 = 5.60x10⁴ (four places until the last digit and third digit is not rounded).
b. 0.00621875 = 6.22x10⁻³ (three places until the first digit and third digit is rounded).
g. 0.003870 = 3.87x10⁻³ (three places until the first digit and third digit is not rounded).
c. 107062.03 = 1.07x10⁵ (five places until the last non-decimal digit and third digit is not rounded).
h. 15.9994 = 16.0 (those consecutive nines round the number to 16.0).
d. 4.00049 = 4.00
i. 12.011 = 12.0
e. 96 = 9.60x10¹ (exponential notation is needed since the number initially has two significant figures so an additional zero is added)
j. 2,000 = 2.00x10³ (three places until the last digit and third digit is not rounded).
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The answer is attached :)
The above question is incomplete, here is the complete question:
Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:
Answer:
The standard molar enthalpy of formation of NO is 90.25 kJ/mol.
Explanation:
To calculate the standard molar enthalpy of formation
...[3]
Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
[1] - [2] = [3]
According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :
=
Answer:
10 g/ml
Explanation:
divide mass by volume means divide 1000 by 100 and your answer will be 10