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pochemuha
3 years ago
15

Atoms in the same PERIOD have the same...

Chemistry
1 answer:
snow_lady [41]3 years ago
3 0

Answer:

A. Number of energy levels

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Ions that are made of more than one atom are examples of
anastassius [24]
These ions are polyatomic ions. The prefix "poly-" means "many," and the rest of the word "atomic" clearly refers to the atoms themselves. So, if there are two or more atoms in an ion, this ion is polyatomic. 
6 0
3 years ago
Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help
fredd [130]

The electronic configuration of Aluminum is 2,8,3.

<h3><u>Explanation:</u></h3>

Aluminum is a metallic element belonging to group 3 of periodic table. Its atomic number is 13.

The rules of electronic configuration is

A. The maximum number of electrons a shell can hold is determined by 2n² where n is the shell number numbered from the one as 1 which is nearest to the nucleus.

B. Each shell has some subshells which actually holds the electrons. Shell 1 has only s subshell where as shell 2 has s and p subshells and so on.

C. Afbau's principle states which subshell of which shell is first filled. For example, 4s subshell is filled before 3d subshell which is stated by Afbau's principle.

The atomic number 13 is filled like- 1st shell has 2 electrons, 2nd shell has 8 electrons and 3rd shell has 3 electrons.

It is also represented as 1s², 2s²,2p⁶, 3s² 3p¹.

6 0
4 years ago
When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?
Temka [501]

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of Mg_3N_2 = 100.0 g

Mass of H_2O = 75.0 g

Molar mass of Mg_3N_2 = 101 g/mol

Molar mass of H_2O = 18 g/mol

First we have to calculate the moles of Mg_3N_2 and H_2O.

\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}

\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol

and,

\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}

\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)

From the balanced reaction we conclude that

As, 6 moles of H_2O react with 1 mole of Mg_3N_2

So, 4.17 moles of H_2O react with \frac{4.17}{6}=0.695 moles of Mg_3N_2

From this we conclude that, Mg_3N_2 is an excess reagent because the given moles are greater than the required moles and H_2O is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 6 moles of H_2O react to give 2 moles of NH_3

So, 4.17 moles of H_2O react to give \frac{2}{6}\times 4.17=1.39 mole of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

Molar mass of NH_3 = 17 g/mole

\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g

Therefore, the maximum theoretical yield of NH_3 is, 23.6 grams.

4 0
3 years ago
49 points~!
Arisa [49]

1. Given the following equation: N2 (g) + 3 H2 (g) ↔ 2NH3 (g) ΔH = -92 kJ/mol

a. this reaction is exothermic as ΔH is -ve

b. the equilibrium will shift 2 the left if nitrogen gas is removed

c. the equilibrium shift 2 the right if the temperature is lowered

d. the equilibrium shift 2 the left if ammonia (NH3) is added

e. principle of thermodynamic potential or Gibbs energy is used to answer B-D

6 0
3 years ago
Read 2 more answers
A hydrocarbon sample was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.00 kg of water rose from 20.
fomenos

Answer:

The heat released by the combustion is 20,47 kJ

Explanation:

Bomb calorimeter is an instrument used to measure the heat of a reaction. The formula is:

Q = C×m×ΔT + Cc×ΔT

Where:

Q is the heat released

C is specific heat of water (4,186kJ/kg°C)

m is mass of water (1,00kg)

ΔT is temperature change (23,65°C - 20,45°C)

And Cc is heat capacity of the calorimeter (2,21kJ/°C)

Replacing these values the heat released by the combustion is:

<em>Q = 20,47 kJ</em>

6 0
3 years ago
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