Answer:
In equation form, Newton's second law of motion is a=Fnetm a = F net m . This is often written in the more familiar form: Fnet = ma. The weight w of an object is defined as the force of gravity acting on an object of mass m. ... Friction is a force that opposes the motion past each other of objects that are touching.
Explanation:
have a nice day :)
Answer:
0.81mole
Explanation:
Given:
Volume of carbon monoxide = 18dm³
Unknown
Number of moles of carbon monoxide = ?
Solution:
To solve this problem;
1 mole of any gas at rtp occupies a volume of 22.4dm³
So;
x mole of carbon monoxide will occupy a volume of 18dm³ at rtp
22.4x = 18
x =
= 0.81mole
The reaction is
Cu⁺²(aq) ---> Cu(s) [reduction]
The species which undergoes reduction is able to oxidize other one thus will called as "oxidizing agent"
Thus Cu⁺² is an oxidizing agent
or Cu is a reducing agent
Al(s) ---> Al⁺³(aq)
Aluminium is undergoing oxidation
Thus is able to reduced others and hence is a reducing agent
Al is a reducing agent
or Al⁺³ is an oxidizing agent
Answer: Cu is the reducing agent, and AI3+ is the oxidizing agent.
Answer:
The value of equilibrium constant is 29.45.
Explanation:
Moles of hydrogen gas = 2.00 mol
Concentration of hydrogen gas =![[H_2]= \frac{2.00 mol}{1.00 L}=2.00 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%20%5Cfrac%7B2.00%20mol%7D%7B1.00%20L%7D%3D2.00%20M)
Moles of iodine gas = 1.00 mol
Concentration of iodine gas =![[I_2]= \frac{I.00 mol}{1.00 L}=1.00 M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%20%5Cfrac%7BI.00%20mol%7D%7B1.00%20L%7D%3D1.00%20M)
![H2(g) + I2(g)\rightleftharpoons 2 HI(g)](https://tex.z-dn.net/?f=H2%28g%29%20%2B%20I2%28g%29%5Crightleftharpoons%202%20HI%28g%29)
initially
2.00 M 1.00 M 1.00 M
At equilibrium:
(2.00-x/2) (1.00-x/2) x
Moles of HI at equilibrium = 1.80 M
Concentration of HI at equilibrium =![[HI]=\frac{1.80 mol}{1.00L} = 1.80M= x](https://tex.z-dn.net/?f=%5BHI%5D%3D%5Cfrac%7B1.80%20mol%7D%7B1.00L%7D%20%3D%201.80M%3D%20x)
The expression of an equilibrium constant is given by ;
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![K_c=\frac{x^2}{(2.00-\frac{x}{2})(1.00-\frac{x}{2})}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bx%5E2%7D%7B%282.00-%5Cfrac%7Bx%7D%7B2%7D%29%281.00-%5Cfrac%7Bx%7D%7B2%7D%29%7D)
Putting x equal to 1.80 M.
![K_c=\frac{(1.80)^2}{(2.00-\frac{1.80}{2})(1.00-\frac{1.80}{2})}=29.45](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%281.80%29%5E2%7D%7B%282.00-%5Cfrac%7B1.80%7D%7B2%7D%29%281.00-%5Cfrac%7B1.80%7D%7B2%7D%29%7D%3D29.45)
The value of equilibrium constant is 29.45.