Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>
<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.
<u>Explanation:</u>
We are given:
Heat of vaporization for water = 2257 J/g
Amount of sweat lost = 307 grams
Applying unitary method:
For 1 g of sweat lost, the energy required is 2257 Joules
So, for 307 grams of sweat lost, the energy required will be = 
Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.
Answer:
2 significant figure is 13.1 aluminium
multiply by 100.4.36×10-5cm
The gravitational force between the objects depends on the mass of the objects and the distance between them.