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Shalnov [3]
3 years ago
13

What is the mass number of an atom with 3 protons and 4 neutrons and 3 electrons

Chemistry
1 answer:
Ksenya-84 [330]3 years ago
7 0
I think its 11.....................
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What is the mass in grams of 1 liter of nitrogen gas?
AlladinOne [14]

Assuming the volume of the gas is measured at standard temperature and pressure, Then one mole of the Gas would occupy 22.4 liters.

Therefore, 1 liter is 1/224 moles

one mole of nitrogen 14 is 14

Therefore 1 liter of the nitrogen weighs 1/224×14

0.0625 grams

7 0
3 years ago
Consider the following reaction at equilibrium: 2 NH3(g) ⇄ N2(g) + 3 H2(g) What does Le Chatelier's principle predict will happe
BabaBlast [244]

Answer : The equilibrium will shift in the left direction.

Explanation :

Le-Chatelier's principle : This principle states that if any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The given reaction is:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

As per question, when we are adding N_2  then the concentration of N_2 is increased on product side then the equilibrium will shift in the direction where decrease of concentration of N_2 takes place. Therefore, the equilibrium will shift in the left direction.

Thus, the equilibrium will shift in the left direction.

7 0
3 years ago
How are Celsius, Kelvin, and Fahrenheit scales similar? How are they different? Considering their differences, think of one scen
PIT_PIT [208]

Answer:

they are similar becauze they measure temperature

6 0
3 years ago
How long does it take the Earth to move from Point A to Point C?
Naddika [18.5K]

Answer:

the answer should be six months

5 0
2 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
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