A mixture of iron and sulfur can be separated by A.magnetic attraction

Which of the compounds below will ionize in

-BARSIC- [3]

I need help on this one

5 0

2 years ago

Write a balanced net ionic equation for the reaction of aqueous solutions of baking soda (NaHCO3) and acetic acid.(A) HCO3–(aq)

IgorC [24]

Answer: The correct answer is Option C.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of sodium bicarbonate and acetic acid is given as:

$NaHCO_3(aq.)+CH_3COOH(aq.)\rightarrow CH_3COONa(aq.)+H_2O(l)+CO_2(g)$

Ionic form of the above equation follows:

$Na^+(aq.)+HCO_3^-(aq.)+CH_3COO^-(aq.)+H^+(aq.)\rightarrow CH_3COO^-(aq.)+Na^+(aq.)+H_2O(l)+CO_2(g)$

As, sodium and acetate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$HCO_3^-(aq.)+H^+(aq.)\rightarrow CO_2(g)+H_2O(l)$

Hence, the correct answer is Option C.

8 0

2 years ago

Most people like to go outside on a sunny day, but too much sun exposure can cause sunburns as well as long term health problems

ss7ja [257]

Earths atmosphere protects you from the sun

4 0

2 years ago

Read 2 more answers

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago

g This plot shows the rate of the decomposition of SO2Cl2 into SO2 and Cl2 as a function of the concentration of SO2Cl2. What is

scoundrel [369]

Answer:

When n = 1, the reaction is of the First Order

Explanation:

Find attach the solution

5 0

2 years ago