CHEMISTRY HIGH SCHOOL

What is the electron structure of nitrogen?1s22s22p3
1s22s23s22p1
1s32s32p1
1s12s12p5

Answers

Answer 1

Answer:

Answer : The correct option is, $1s^22s^22p3$

Explanation :

Electronic configuration : It is defined as the arrangement of the electrons of an atom in an atomic orbital.

The given element is, nitrogen.

The nitrogen element belongs to group 15 and the atomic number of nitrogen is 7.

As we know that the number of electrons is equal to the number of atomic number.

The number of electrons present in nitrogen = 7

Thus, the electronic configuration of nitrogen is, $1s^22s^22p3$

Answer 2

Answer:

1s22s22p3 considering N is the third element in the 2nd P sublevel

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Consider a neatly packed box of marbles compared to what you get if you dump the marbles on the floor. Which state has the higher entropy. a. spilled marbles b. marbles in packed box c. none d. both

Answers

Answer:

a. spilled marbles

Explanation:

entropy is the tendency toward a state of disorder or randomness, so the spilled box of marbles has the higher entropy

Final answer:

The spilled marbles and marbles in a packed box have different states of entropy.

Explanation:

Entropy is a concept in thermodynamics that refers to the disorder or randomness of a system. In this case, the state with the higher entropy would be the spilled marbles. When marbles are neatly packed in a box, they are in a more ordered and organized state, resulting in lower entropy. However, when the marbles are spilled on the floor, they are now in a more disordered and random state, leading to higher entropy. Therefore, the answer is d. both.

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What is a negatively charged subatomic particle?

Answers

i think its electrons

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water (d = 1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50 % v/v (b) mass percent 52.7 % w/w (c) molarity M (d) molality m (e) mole fraction

Answers

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

$\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}* \text{Volume of ethylene glycol}=1.114g/mL* 1mL=1.114g$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1.00g/mL* 1mL=1.00g$

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(2.114g)/(1.070g/mL)=1.975mL$

(a) Now we have to calculate the volume percent.

$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}* 100=(1mL)/(1.975mL)* 100=50.63\%$

(b) Now we have to calculate the mass percent.

$\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}* 100=(1.114g)/(2.114g)* 100=52.69\%$

(c) Now we have to calculate the molarity.

$\text{Molarity}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Volume of solution (in mL)}}$

$\text{Molarity}=(1.114g* 1000)/(62.07g/mole* 1.975L)=9.087mole/L$

(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Mass of water (in g)}}$

$\text{Molality}=(1.114g* 1000)/(62.07g/mole* 1kg)=17.947mole/kg$

(e) Now we have to calculate the mole fraction of ethylene glycol.

$\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}$

$\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=(1.114g)/(62.07g/mole)=0.01795mole$

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=(1g)/(18g/mole)=0.0555mole$

$\text{Mole fraction of ethylene glycol}=(0.01795mole)/(0.01795mole+0.0555mole)=0.244$

How many moles of CaO are needed to react with an excess of water to form 370 grams of calcium hydroxide?

Answers

The required moles of CaO are needed to react with an excess of water to form 370 grams of calcium hydroxide is 280g.

What is the relation between mass & moles?

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

W = given mass
M = molar mass

Given chemical reaction is:

CaO + H₂O → Ca(OH)₂

Moles of Ca(OH)₂ = 370g / 74g/mol = 5mol

From the stoichiometry of the reaction, 5 moles of CaO is required to produce 5 moles of Ca(OH)₂.

Mass of CaO = (5mol)(56g/mol) = 280g.

Hence required mass of CaO is 280g.

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CaO + H20 --> Ca(OH)2

Calcium Hydroxide has a relative molecular mass of approximately 74 grams. Therefore, with 370 grams provided, we have 370g/74g, or 5 moles.

Therefore, due to molar ratios, 5 moles of CaO must have reacted with 5 moles of H2O to form 5 moles of Ca(OH)2.

The answer to your question is 5 moles of CaO or 280 g of CaO

What is the atomic mass of w

Answers

Answer:

183.84 u

Explanation:

W (also known as "Tungsten") atomic number is 74 and atomic mass is 183.84 u

Which chemical equation is correctly balanced?(1) H2(g) + O2(g)=> H2O(g)
(2) N2(g) + H2(g) =>NH3(g)
(3) 2NaCl(s)=>Na(s) + Cl2(g)
(4) 2KCl(s) => 2K(s) + Cl2(g)

Answers

2KCl(s) => 2K(s) + Cl2(g) is correctly balanced. Becuase 2 K, 2 Cl on the left, and 2 K, 2 Cl on the right.

The correct balanced chemicalequation is 2KCl(s) => 2K(s) + Cl2(g).

The correct balanced chemical equation from the options provided is (4) 2KCl(s) => 2K(s) + Cl2(g).

In this equation, two molecules of potassium chloride (KCl) react to form two molecules of potassium (K) and one molecule of chlorine gas (Cl2). This equation is balanced because the number of atoms of each element is equal on both sides of the equation.

For example, there are two atoms of potassium and two atoms of chlorine on both sides of the equation, ensuring the equation is balanced.

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