Answer:
92.72 kJ
Explanation:
2 N₂ (g) + O₂ (g) —-> 2 N₂O
According to question , one mole of N₂O requires 163.2 kJ of heat
Molecular weight of N₂O = 44 gm
25 g N₂O = 25 / 44 mole
25 / 44 mole will require 163.2 x 25 / 44 kJ
= 92.72 kJ
Answer:
Rate depends on the rate constant. The rate constant depends on temperature and activation energy. If you have lower activation energy the rate will be higher. This is why catalysts are added since catalysts provide an alternate pathway that requires lower activation energy and catalysts are added to increase the rate of reaction.
Explanation:
This is only the answer if you were asking:
"Which corresponds to the faster rate: a mechanism with a small activation energy or one with a large activation energy?"
Thats what I understood about your question.
ANWERS ~
We know that :
1 cal (th) = 4.184 J
1 J = 0.2390057361 cal (th) , so :
•55.2 j to cal > 13.193116635 cal
•110 call > 460.24 joule
•65 kj > divide the energy value by 4.184
= 15.535 kilocalories calorie (IT)
——————
Converting form C to F > (F-32)*5/9Understand it better if we have Fahrenheit just add to the equation mentioned to find Celsius.
+to find F to C> (9/5*C)+32
•425 Fahrenheit = (425- 32) × 5/9 =218.33333333 Celsius
•1935 C = 3515 F
———————————-
Converting Celsius to kelvin,We know that :
K = C + 273.15
C = K - 273.15
And from F to K=9/5(F+459.67)
And K to F =(9/5 *k)-459.67
•39.4 Celsius = 312.55 kelvin
•337 Fahrenheit = (337+ 459.67) × 5/9 =442.594 kelvin
Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>
The process of preparing solutions from stock solutions of higher concentration is known as dilution.
Dilution is done with the aid of the dilution formula given below:
where
- C1 is the concentration of stock solution
- V1 is the volume of stock solution required to prepare a diluted solution
- C2 is the concentration of the diluted solution prepared
- V2 is the final volume of the diluted solution
From the data provided:
C1 is not given
V1 is unknown
C2 = 25%
V2 = 12 mL
- Assuming C1 is 50% solution
Volume of stock, V1, required is calculated as follows:
V1 = C2V2/C1
V1 = 25 × 12 /50
V1 = 6 mL
Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
Learn more about dilution formula at: brainly.com/question/7208546
