Answer:
Waves carry energy from one place to another.
Because waves carry energy,some waves are used for communication,eg radio and television waves and mobile telephone signals.
Explanation:
i hope it helps
that's my answer
correct me if im wrong
<h3>#
<em><u>c</u></em><em><u>a</u></em><em><u>r</u></em><em><u>r</u></em><em><u>y</u></em><em><u>o</u></em><em><u>n</u></em><em><u>l</u></em><em><u>e</u></em><em><u>a</u></em><em><u>r</u></em><em><u>n</u></em><em><u>i</u></em><em><u>n</u></em><em><u>g</u></em></h3>
<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.
</span><span>E] MnO2(s) MnO4-(aq</span>
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is ![K_c = 14.39](https://tex.z-dn.net/?f=K_c%20%20%3D%2014.39)
Explanation:
The chemical equation for this decomposition of ammonia is
↔ ![N_2 + 3 H_2](https://tex.z-dn.net/?f=N_2%20%2B%203%20H_2)
The initial concentration of ammonia is mathematically represented a
![[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%20%20%5Cfrac%7Bn_1%7D%7BV_1%7D%20%20%3D%20%5Cfrac%7B29%7D%7B75%7D)
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
The initial concentration of nitrogen gas is mathematically represented a
![[N_2] = \frac{n_2}{V_2}](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%20%5Cfrac%7Bn_2%7D%7BV_2%7D)
![[N_2] = 0.173 \ M](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%200.173%20%20%5C%20%20M)
So looking at the equation
Initially (Before reaction)
![N_2 = 0 \ M](https://tex.z-dn.net/?f=N_2%20%20%3D%20%200%20%5C%20%20M)
![H_2 = 0 \ M](https://tex.z-dn.net/?f=H_2%20%3D%20%200%20%5C%20M)
During reaction(this is gotten from the reaction equation )
(this implies that it losses two moles of concentration )
(this implies that it gains 1 moles)
(this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
![N_2 = x](https://tex.z-dn.net/?f=N_2%20%3D%20%20x)
![H_2 = 3 x](https://tex.z-dn.net/?f=H_2%20%20%3D%20%203%20x)
Now since
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
Now the equilibrium constant is
![K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=K_c%20%20%3D%20%20%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
substituting values
![K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}](https://tex.z-dn.net/?f=K_c%20%20%3D%20%20%5Cfrac%7B%280.173%29%20%280.519%29%5E3%7D%7B%280.041%29%5E2%7D)
![K_c = 14.39](https://tex.z-dn.net/?f=K_c%20%20%3D%2014.39)
Answer:
the answer should be henry's law