Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
Answer:
45.95 Jkg^-1°C^-1
Explanation:
as specific heat capacity = heat energy / mass × delta
temperature
=52500/10.2×112
=45.95 Jkg^-1°C^-1
Answer:
Explanation:
To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.
The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.
This can be expressed below:
RAM = Σmₙαₙ
where mₙ is the mass of isotope n
αₙ is the abundance of isotope n
for this problem:
RAM of Li = m₆α₆ + m₇α₇
m₆ is mass of isotope Li-6
α₆ is the abundance of isotope Li-6
m₇ is mass of isotope Li-7
α₇ is the abundance of isotope Li-7
That will make a gold-202 nucleus.
<h3>Explanation</h3>
Refer to a periodic table. The atomic number of mercury Hg is 80.
Step One: Bombard the
with a neutron
. The neutron will add 1 to the mass number 202 of
. However, the atomic number will stay the same.
- New mass number: 202 + 1 = 203.
- Atomic number is still 80.
.
Double check the equation:
- Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.
Step Two: The
nucleus loses a proton
. Both the mass number 203 and the atomic number will decrease by 1.
- New mass number: 203 - 1 = 202.
- New atomic number: 80 - 1 = 79.
Refer to a periodic table. What's the element with atomic number 79? Gold Au.
.
Double check the equation:
- Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.
A gold-202 nucleus is formed.