Pls help !! it’s a timed quiz answer

Electrolytic cells use electricity to cause a nonspontaneous redox reaction to occur. An electrolytic cell is constructed using

Fittoniya [83]

Answer:

At anode - ${Br_2}_{(l)}$

At cathode - ${H_2}_{(g)}$

Explanation:

Electrolysis of NaBr:

Water will exist as:

$H_2O\rightleftharpoons H^++OH^-$

The salt, NaBr will dissociate as:

$NaBr\rightarrow Na^++Br^-$

At the anode, oxidation takes place, as shown below.

${Br^-}_{(aq)}\rightarrow {Br_2}_{(l)}+2e^-$

At the cathode, reduction takes place, as shown below.

$2{H^+}_{(aq)}+2e^-\rightarrow {H_2}_{(g)}$

4 0

3 years ago

What is the correct balanced equation for the reaction of sodium with water?

Wittaler [7]

Answer:

A. 2Na + 2H₂O → 2NaOH + H₂

Explanation:

When sodium is reacted with water, a single replacement reaction occurs. The product of the reaction is typically sodium hydroxide and hydrogen gas.

Reactants:

Sodium + water

Product:

Sodium hydroxide + hydrogen gas

So;

2Na + 2H₂O → 2NaOH + H₂

6 0

2 years ago

How do I solve this problem?

JulijaS [17]

First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps

5 0

2 years ago

The element in Group 14 Period 4 has how many neutrons and is it a metal, nonmetal, metalloid or noble gas.

monitta

Number of neutrons=41, and it is metalloid

<h3>Further explanation</h3>

Group 14⇒valence electron = 4(ns²np²)

Period 4⇒n=4

So electron configuration of the element :

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p² = 32 electron=atomic number

The element with atomic number 32, which is in period 4 and group 14 is Ge-Germanium

There are seven elements that can be classified as metalloids, namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po).

While the mass number: 73

So the number of neutrons = mass number-atomic number

$\tt 73-32=41$

8 0

2 years ago

Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa

Lapatulllka [165]

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason:

If you look at the combining masses of X and Y in each of the two compounds,

The first compound contains 0.25g of X combined with 0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3

The second compound contains 0.33 g of X combined with 0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!)

The first compound will then contain 4g of X and 12g of Y
The second compound will then contain 6g of X and 12g of Y

<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

8 0

2 years ago