The plants give out oxygen so for us humans will live
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.
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Grams of Ca(NO₃)₂ produced : 0.985 g
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction
CaCl₂ + 2AgNO₃ → 2AgCl + Ca(NO₃)₂
MW AgNO₃ : 107.9+14+3.16=169.9
mol AgNO₃ :
mol ratio Ca(NO₃)₂ : AgNO₃ = 1 : 2, so mol Ca(NO₃)₂ :
MW Ca(NO₃)₂ : 40.1+2.14+6.16=164.1 g/mol
mass Ca(NO₃)₂ :
When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
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