Given:
Iron, 125 grams
T
1 = 23.5 degrees Celsius, T2 =
78 degrees Celsius.
Required:
Heat produced in kilojoules
Solution:
The molar mass of iron is 55.8
grams per mole. SO we need to change the given mass of iron into moles.
Number of moles of iron = 125 g/(55.8
g/mol) = 2.24 moles
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314
Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or
1.015 kilojoules</u>
<span>This is the amount of heat
produced in warming 125 g f iron.</span>
Cmasdfghjkl;^2 I think
Lol hoped this helped
A. Fireworks exploding is the answer
Answer:
Ca(NO3)2 has the highest boiling point ( option A)
Explanation:
Step 1: Data given
A. 1.25 M Ca(NO3)2
B. 1.25 M KNO3
C. 1.25 M CH3OH
D. 2.50 M C6H12O6
Step 2: Calculate highest boiling point
The boiling point depends on the van't Hoff factor
This shows the particles produced when the substance is dissolved. For non-electrolytes dissolved in water, the van' t Hoff factor is 1.
Ca(NO3)2 → Ca^2+ + 2NO3- → Van't Hoff factor = 3
KNO3 → K+ + NO3- → Van't Hoff factor = 2
CH3OH is a non-elektrolyte → Van't Hoff factor = 1
C6H12O6 is a non-elektrolyte → Van't Hoff factor = 1
Ca(NO3)2 has the highest boiling point
Answer:
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)
Explanation:
HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.
. V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)](https://tex.z-dn.net/?f=n_%7BH%5E%7B%2B%7D%20%7D%20from%20HCl%20%3D%20%5BHCl%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHCl%7D%28L%29%20%20%5C%5C%20n_%7BH%5E%7B%2B%7D%20%7D%20from%20HNO_%7B3%7D%20%20%3D%20%5BHNO_%7B3%7D%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHNO_%7B3%7D%7D%28L%29)
Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows
For molar concentration of hydrogen ions:
![[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%20%3D%20%5Cfrac%7Bn_%7BH%5E%7B%2B%7D%7D%28mol%29%7D%7BV%28L%29%7D)
![[H^{+}] = \frac{0.761}{1.00}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B0.761%7D%7B1.00%7D)
From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows
![K_{w} = [H^{+} ][OH^{-} ]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%20%5D)
![[OH^{-}]=\frac{Kw}{[H^{+}] }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7BKw%7D%7B%5BH%5E%7B%2B%7D%5D%20%7D)
![[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7B1.01X10-%5E%7B-14%7D%7D%7B0.761%20%7D)
The pH of the solution can be measured by the following formula:
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)
