Answer:
96.09 g/mol
Explanation:
You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.
If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).
Also, please take note that I will be using the unit g/mol for all the weights. Thus,
Step 1
N = 14.01 g/mol
H = 1.008 g/mol
O = 16.00 g/mol
C = 12.01 g/mol
Since your compound is
(
N
H
4
)
2
C
O
3
, you need to multiply the atomic weights by their subscripts. Therefore,
Step 2
N = 14.01 g/mol × 2 =
28.02 g/mol
H = 1.008 g/mol × (4×2) =
8.064 g/mol
O = 16.00 g/mol × 3 =
48.00 g/mol
C = 12.01 g/mol × 1 =
12.00 g/mol
To get the mass of the substance, we need to add all the weights from Step 2.
Step 3
molar mass of
(
NH
4
)
2
CO
3
=
(28.02 + 8.064 + 48.00 + 12.01) g/mol
=
96.09 g/mol
this is a google search and a example i hope is helps to solve
Answer:
d. 10
Explanation:
The 24 means the atomic mass and twelve means the number of electrons. You subtract the number of electrons from the atomic mass. 24-12=12
Then, since there's +2, you subtract 2 from 12 which equals 10 which means there are 10 electrons now.
The confusing thing is that + means subtract and - means add here.
Hope this helps.
<span>iron has 26 protons and 30 neutrons are in the nucleus</span>
Answer: B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O (g) (ΔH = -2035 kJ/mol) 3H2O (g) → 3H2O (l) (ΔH = -132 kJ/mol) 3H2O (l) → 3H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)
Explanation: ??
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
