Answer:
The empirical formula is SF6 (option E)
Explanation:
Step 1: Data given
Mass of sulfur = 3.21 grams
Mass of fluorine = 11.4 grams
Molar mass sulfur = 32.065 g/mol
Molar mass fluorine = 19.00 g/mol
Step 2: Calculate moles
Moles = mass /molar mass
Moles sulfur = 3.21 grams / 32.065 g/mol
Moles sulfur = 0.100 moles
Moles fluorine = 11.4 grams / 19.00 g/mol
Moles fluorine = 0.600 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
S: 0.100 / 0.100 = 1
F : 0.600 / 0.100 = 6
The empirical formula is SF6 (option E)
Answer:
in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution. (option C).
Explanation:
In a reaction where NaOH is added to H2O.
NaOH is considered a strong base, this means that in an aqueous solution ( in water) it's able to completely disassociate in ions.
There will not remain any NaOH in the solution. This means option D is not correct.
The ions in which NaOH will disassociate are : NaOH → Na+ + OH-
These ions we will find in the solution.
Not only Na+ because NaOH is a strong base, so there will be a lot of OH- ions as well in solution.
This means in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution.
Answer is:<span>the yield is 50%.
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Chemical reaction: C + O₂ → CO₂.
n(C) = 0.3 mol; amount of substance.
n(O₂) = 0.3 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
n(CO₂) = 0.3 mol.
M(CO₂) = 44 g/mol; molar mass of caron(IV) oxide.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) =0.3 mol · 44 g/mol.
m(CO₂) = 13.2 g; mass of carbon(IV) oxide.
the yield = 6.6 g ÷ 13.2 g · 100%.
the yield = 50%.
Answer:
C. at low temperature and low pressure.
Explanation:
- <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
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<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>
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<em><u>Effect of pressure:</u></em>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.
<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>
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<u><em>Effect of temperature:</em></u>
- The reaction is exothermic because the sign of ΔH is (negative).
- So, we can write the reaction as:
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>
- Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>
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<em>C. at low temperature and low pressure.</em>
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