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frez [133]
3 years ago
8

The electronic structures of atoms X and Y are shown.

Chemistry
1 answer:
Temka [501]3 years ago
5 0

Answer:

Its formula is B) XY3,also known as ammonia,because X is hydrogen and Y is nitrogen.

So,yeah the answer is B

Pls mark me as brainliest

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What is the molarity of a solution that contains 0.400 mol HCI in 9.79 L solutions ​
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Molarity's formula is known as: Molarity(M)=moles of solute/liters solution.

In this case we are already given moles and liters so you just have to plug the numbers into the equation.

0.400 mol HCL/9.79L solution=0.040858M

If you were to use scientific notation, the answer will be: 4.1*10^-2, but otherwise, you can just use the decimals above and round appropriately as you see fit.

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Which example is the site of reduction when the dry cell is operating
Simora [160]

Answer:

A battery contains electrochemical cells that can store chemical energy to be converted to electrical energy. A dry-cell battery stores energy in an immobilized electrolyte paste, which minimizes the need for water. Common examples of dry-cell batteries include zinc-carbon batteries and alkaline batteries.

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In an electrically neutral atom of any element, there are equal numbers of
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3 years ago
What kind of questions CANNOT be answered by chemistry?
Sergeu [11.5K]

Answer:

d. why matter exists

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The kind of questions that chemistry CANNOT answer is "why matter exists".

In Chemistry, question of how the properties, composition and structure of substances are is answered. Also, the transformations that these substances undergo, and the energy that they release or absorbe during the transformation processes are revealed in chemistry.

Chemistry can answer the question of what forms of matter exists but cannot answer why matter actually exists.

5 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

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[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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