The time it will take for a car travelling 88 km/hr (55mi/hr) to travel 500km would be 3.5 hours.
How? Add 88 + 58.
That'll equal 143, then divide 500 ÷ 143 = 3.5
So, 3.5 is the time that the car will take travelling 88 km/hr trying to reach the destination of 500km.
Answer:
b- 4.4 * 10^-12.
Explanation:
Hello.
In this case, as the reaction:
A + 2B → 3C
Has an equilibrium expression of:
![K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BC%5D%5E3%7D%7B%5BA%5D%5BB%5D%5E2%7D%3D2.1x10%5E%7B-6%7D)
If we analyze the reaction:
2A + 4B → 6C
Which is twice the initial one, the equilibrium expression is:
![K_2=\frac{[C]^6}{[A]^2[B]^4}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BC%5D%5E6%7D%7B%5BA%5D%5E2%5BB%5D%5E4%7D)
It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:
Thus, the equilibrium constant of the second reaction turns out:
Therefore, the answer is b- 4.4 * 10^-12.
Best regards.
Answer: W + BgCz2 arrow WCz + Bg
2 W + BgCz2 arrow 2 WCz + Bg
Explanation:
Cz has 2 so you balcne the other side of WCz.
Since you Balcanes the Cz you changed the W and you Balcanes the other W on the left side.
Answer:
7.32 g of F₂
Solution:
The equation is as follow,
2 LiI + F₂ → 2 LiF + I₂
According to equation,
51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂
Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g
X = 7.32 g of F₂
