Answer:
0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.
Ans: a blood protein produced in response to and counteracting a specific antigen. Antibodies combine chemically with substances which the body recognizes as alien, such as bacteria, viruses, and foreign substances in the blood.
Kinetic and nuclear are energies
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
Answer:
Strike slip forms at transform boundaries
