Answer:
2H₂ + O₂ → 2H₂O
Explanation:
Chemical equation:
H₂ + O₂ → H₂O
Balance chemical equation:
2H₂ + O₂ → 2H₂O
Step 1:
H₂ + O₂ → H₂O
Left hand side Right hand side
H = 2 H = 2
O = 2 O = 1
Step 2:
H₂ + O₂ → 2H₂O
Left hand side Right hand side
H = 2 H = 4
O = 2 O = 2
Step 3:
2H₂ + O₂ → 2H₂O
Left hand side Right hand side
H = 4 H = 4
O = 2 O = 2
The balanced chemical reaction is:
<span>C4H8O2 + 2H2--> 2C2H6O
</span>A. How many grams of ethyl alcohol are produced by reaction of 2.7 mol of ethyl acetate with H2? 2.7 mol C4H8O2 ( 2 mol C2H6O / 1 mol C4H8O2) (46.07 g / 1 mol) = 248.78 g ethyl alcohol
B. How many grams of ethyl alcohol are produced by reaction of 13.0g of ethyl acetate with H2?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol C2H6O / 1 mol C4H8O2) (<span>46.07 g / 1 mol) = 13.59 g ethyl alcohol</span>
C. How many grams of H2 are needed to react with 13.0g of ethyl acetate?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol H2 / 1 mol C4H8O2) (2.02 g / 1 mol) = 0.5961 g H2
Given :
Length , l = 59.8 m.
Breadth , b = 26.6 m.
Depth , d = 3.7 ft .
Density of water , 
.
To Find :
Mass of water in pool .
Solution :
First we will covert depth into m from ft .
For ,
So , volume of pool is :
We know , density is given by :
So , 
Putting given values in above equation :
Hence , this is the required solution.
The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 
10^-5 V.
<u>Explanation:</u>
Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.
Given p = 1.47 D = 1.47 3.34 10^-30 = 4.90 10^-30.
V = 1 / (4π∈о) (p cos(θ)) / (r^2)
where p is a permanent electric dipole,
∈ο is permittivity,
r is the radius from the axis of a dipole,
V is the electric potential.
V = 1 / (4 3.14 8.85 10^-12) (4.90 10^-30 1) / (55.3 10^-9)^2
V = 1.44 10^-5 V.
