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3 years ago

Write the name of the first twenty elements with their symbols and their atomic number

Chemistry

2 answers:

7nadin3 [17]3 years ago

6 0

Answer:

1 Hydrogen H

2 Helium He

3 Lithium Li

4 Beryllium Be

5 Boron B

6 Carbon C

7 Nitrogen N

8 Oxygen O

9 Fluorine F

10 Neon Ne

11 Sodium Na

12 Magnesium Mg

13 Aluminium Al

14 Silicon Si

15 Phosphorus P

16 Sulfur S

17 Chlorine Cl

18 Argon Ar

19 Potassium K

20. calcium Ca

Archy [21]3 years ago

3 0

Answer:

h - hydrogen (1)

he - helium (2)

li - lithium (3)

be - beryllium (4)

b - boron (5)

c - carbon (6)

n - nitrogen (7)

o - oxygen (8)

f - fluorine (9)

ne - neon (10)

na - sodium (11)

mg - magnesium (12)

al- aluminium (13)

si - silicon (14)

p - phosphorus (15)

s - sulfur (16)

cl - chlorine (17)

ar - argon (18)

k - potassium (19)

ca - calcium (20)

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Answer:

C. 11 moles of N2O

Explanation:

A. CO2 exists as a molecular compound. The number of atoms present = 1 atom of Carbon and 2 atoms oxygen = 3 atoms * 9 = 18 atoms

B. Xe exists as an atom. Number of atoms present = 10 * 1 atom = 10 atoms

C. N20 contains 3 atoms; 2 atoms of Nitrogen and 1 atom of oxygen.

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3 years ago

How would you abbreviate you as an element?

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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ 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, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

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

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0

3 years ago

Write the name of the first twenty elements with their symbols and their atomic number​

Write the name of the first twenty elements with their symbols and their atomic number