Answer:
0.090 J/(mmol·°C) × (1000 mmol/mole × 1 kJ/(1000 J)) = 0.090 kJ/mole
Explanation:
The unit of conversion from kilo-Joules to Joules is given as follows;
1000 Joules = 1 kilo-Joule
The unit of conversion from milimoles to moles is given as follows;
1000 milimoles = 1 Mole
Therefore, we have
The value of the given expression is 0.090 J/(mmol·°C) × 1000 mmol/mole × 1 kJ/(1000 J) = 0.090 kJ/mole
0.090 J/milimole = 0.09 kJ/mole.
To write quantities in ordinary notation, you need to notice the power of the exponent.
If the power is positive, then you move the decimal point to the right by the number the power in exponent tells you.
If the power is negative, then you move the decimal point to the left by the number the power in exponent tells you.
We have:
3 x 10^-4 : the power of exponent is negative, therefore, we will move the decimal point 4 places to the left.
3 x 10^-4 = 0.0003 km
3 x 10^4 : he power of exponent is positive, therefore, we will move the decimal point 4 places to the right.
3 x 10^4 = 30000 km
In the periodic table the lanthanoid and the actinides are place separately at the bottom because of their electronic configuration and their properties compared to the other elements.
The the lanthanoid and the actinides are place separately at the bottom in the periodic table due to their electronic configuration and the properties. and to make the periodic table more convenient . if we place the f block elements that is he lanthanoid and the actinides then the size of the periodic table will increase. the f block elements are called as the inner transition element.
Thus , to make the periodic table more convenient and to group the elements in the block the the lanthanoid and the actinides are place separately at the bottom.
To learn more about lanthanoid and actinide here
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Answer:
31.9 °C
Explanation:
The formula for the heat q absorbed by an object is
q = mCΔT where ΔT = (T₂ - T₁)
Data:
q = 12.35 cal
m = 19.75 g
C = 0.125 cal°C⁻¹g⁻¹
T₂ = 37.0 °C
Calculations
(a) Calculate ΔT
q = mCΔT
12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT
12.35 = 2.406ΔT °C⁻¹
ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C
(b) Calculate T₂
ΔT = T₂ - T₁
T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C
The original temperature was 31.9 °C.
