The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)
find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles
since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles
by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles
= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen
I believe its a. compound<span />
We have that
The average atomic mass round and to the second decimal place (0.01) is
A_m=126.90
From the question we are told
Find the average atomic mass round and to the second decimal place (0.01)
80% 127I, 17% 126I, 3% 128I
Generally the equation for the Average Mass is mathematically given as
Where
a1 = 127
b1 = 80%
a2 = 126
b2 = 17%
a3 = 128
b3 = 3%
Therefore
In conclusion
The average atomic mass round and to the second decimal place (0.01) is
A_m=126.90
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<h3>
Answer:</h3>
6AgI + Fe₂(CO₃)3 → 2FeI₃+ 3Ag₂CO₃
<h3>
Explanation:</h3>
- The process of balancing chemical equations involves putting coefficients on reactants and products to ensure the number of atoms of each element is equal on both sides of the equation.
- It is a try and error method that is done to make an equation obey the law of conservation of mass.
In this case;
- Our unbalanced equation is, AgI + Fe₂(CO₃)3 → FeI₃+ Ag₂CO₃
- It can be balanced by putting the coefficients 6, 1, 2, 3.
- Therefore, the balanced chemical equation is;
6AgI + Fe₂(CO₃)3 → 2FeI₃+ 3Ag₂CO₃