Mass of oxygen : 54 g
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
Reaction
mass before reaction = mass after reaction
mass C + mass O₂= mass CO₂
b)They have different numbers of neutrons and different charges.
c)They have different numbers of protons and different mass numbers.
d)They have different numbers of neutrons and different mass numbers.
Boron occurs naturally as two isotopes that is they have the same number of protons and different numbers of neutrons with different atomic charges. Option D is correct.
These are the variations of elements having the same number of protons and electrons and different numbers of neutrons. As they are neutral in charge and their mass is totally negligible as compared to the mass of proton as they both are located at the center of the atom
The center of the atom is the nucleusas there are some more species of element isobars isotones isoelectronic. Isobars are those species that have the same neutronic numbers with different photonic and electronic numbers
The isotopes of boron are boron10, boron12, boron14, etc. .Carbon is another compound that is having 3 isotopes carbon12 carbon13, and caobon14. But mostly their properties are the same with a little difference between them
Therefore, the correct option is D.
Learn more about isotopes, here:
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Answer:
P₂ = 2.19 atm
Explanation:
Given data:
Initial volume = 1.5 L
Initial pressure =1 atm
Initial temperature = 273K
Final temperature = 26°C (26+273 = 299 K)
Final volume = 0.75 L
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 1 atm × 1.5 L × 299 K / 273 K × 0.75 L
P₂ = 448.5 atm .L. K / 204.75 K.L
P₂ = 2.19 atm
Answer : The concentration of NaOH is, 0.336 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Thus, the concentration of NaOH is, 0.336 M