Answer:
Benzene
Explanation:
You have to find the freezing point for each of these (make sure they're in Celsius). You might want to double check these but:
Benzene: 5.5
Water: 0
Butane: -138
Nitrogen: -210
When we balance the given equation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We will get
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
Solution:
Balancing the given equaation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We have to balance the given number of O
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
We get balanced equation
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
The reaction quotient will be
Qc = [product] / [reactant]
Qc = [SO₂F₂(g)] / [SF₆(g) + SO₃(g)]
To learn more click the given link
brainly.com/question/16025432
#SPJ4
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:
C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
Answer is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂) = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.
B, As only are metalloids or have semimetallic properties.
