The ionic formula of sodium oxide would be Na20
Answer:
D.
Explanation:
Hello,
In this case, the isomer of an organic compound is another organic compound having the same molecular formula but different structural formula, thus, the given compound's molecular formula is C₅H₈ since it is an alkyne due to the triple bond. Next, we analyze each option:
A. C₅H₁₂
B. C₅H₁₀
C. C₅H₁₀
D. C₅H₈
For that reason answer is D. based on the molecular formula as well as due to the presence of the triple bond unsaturation (alkyne as well).
Best regards.
H₂S
<h3>Further explanation</h3>
Given
ΔH fusion and ΔH vaporization of different substances
Required
The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes
Solution
We can use the formula :
Q=heat/energy absorbed
n = moles
The heat absorbed : 58.16 kJ
moles = 3.11
so ΔH vaporization :
The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S
(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)
Answer:
Mass of NH₃ produced = 34 g
Explanation:
Given data:
Mass of nitrogen = 28 g
Mass of Hydrogen = 12 g
Mass of NH₃ produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 28 g/ 28 g/mol
Number of moles = 1 mol
Moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 12 g/ 2 g/mol
Number of moles = 6 mol
Now we will compare the moles of hydrogen and nitrogen with ammonia.
H₂ : NH₃
3 : 2
6 : 2/3×6 = 4 mol
N₂ : NH₃
1 : 2
Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.
Mass of ammonia produced:
Mass = number of moles × molar mass
Mass = 2 mol × 17 g/mol
Mass = 34 g
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
