Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as
Where 
is the quantum number
now
We can look at the given angular momentum in this form as
comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
Answer:
184.113 g/mol
Explanation: The atomic mass of Mg is 24.3 amu. The atomic mass of bromine is 79.9. Therefore, the formula weight of MgBr2 equals 24.3 amu + (2 × 79.9 amu), or 184.1 amu. Because a substance's molar mass has the same numerical value as its formula weight, the molar mass of MgBr2 equals 184.1 g/mol.
The answer is 0.5010
Number of moles (n) is equal to the quotient of mass (m) and molar mass of a sample (Mr):
n = m/Mr
We have:
n = ?
m = 40.10 g
Mr = 80.0432 g/mol
n = 40.10 g : 80.0432 g/mol = 0.5010
<span>40.10 has 4 significant digits,
</span><span>80.0432 has 6 significant digits.
Since 4 is less than 6, we choose 4 </span>significant digits
Answer:
10.80
Explanation:
As per the equation, let us calculate the mole ratio. N2+3H2→2NH3. As per the equation one mole of nitrogen reacts with 1 mol of hydrogen.
In terms of mass. 28.01 g of nitrogen needs 3 mol of hydrogen or 6.048 g of hydrogen.
We can set up the ratio;
28.01 g of
l
N
2
needs
6.048 g of
l
H
2
1 g of
l
N
2
needs
6.048
28.01
g of
l
H
2
50.0 g of
l
N
2
needs
6.048
×
50.0
28.01
l
g of
l
H
2
=
10.80 g of
l
H
2
The answer would be A. troposphere
