The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.
<h3>How do we calculate atomic mass?</h3>
Atomic mass (A) of any atom will be calculated as:
A = mass of protons + mass of neutrons
In the Thorium-234:
Number of protons = 90
Number of neutrons = 144
Mass of one proton = 1.00728 amu
Mass of one neutron = 1.00866 amu
Mass of thorium-234 = 90(1.00728) + 144(1.00866)
Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu
Given mass of thorium-234 = 234.04360 amu
Mass defect = 235.90224 - 234.04360 = 1.85864 amu
Hence required value is 1.85864 amu.
To know more about Atomic mass (A), visit the below link:
brainly.com/question/801533
Answer:
54g of water
Explanation:
Based on the reaction, 1 mole of methane produce 2 moles of water.
To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:
<em>Molar mass CH₄:</em>
1C = 12g/mol*1
4H = 1g/mol*4
12g/mol + 4g/mol = 16g/mol
<em>Moles methane: </em>
24g CH₄ * (1mol / 16g) = 1.5 moles methane
<em>Moles water:</em>
1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O
<em>Molar mass water:</em>
2H = 1g/mol*2
1O = 16g/mol*1
2g/mol + 16g/mol = 18g/mol
<em>Mass water:</em>
3.0moles H₂O * (18g / mol) =
<h3>54g of water</h3>
The volume did not change, it remained at 20 ml
<h3>Further explanation</h3>
Given
20 ml a sample gas at STP(273 K, 1 atm)
T₂=546 K
P₂=2 atm
Required
The volume
Solution
Combined gas Law :
Input the value :
The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)
Answer: Option (d) is the correct answer.
Explanation:
According to Bronsted-Lowry, species which donate a proton are known as acid. The species which accept a proton are known as a base.
In the given reaction, acids and bases are as follows.
HI + +
Acid Base Conjugate acid Conjugate base
Therefore, the acid HI loses a proton to form a conjugate base that is .
Thus, we can conclude that HI and is an acid conjugate base pair.