When an aluminum cube of edge 0.100 m and temperature T0= 5.00°C is submerged in a bucket of water, the level of the water rises
1 answer:
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Answer:
1) 2.0 g
2) 0 g
3) 4.17 g
4) 2.57 g
Explanation:
First of all, we need to know the compounds and the reaction. The ion carbonate is , and the ion nitrate is
.
Sodium is in group 1, so it must lose one electron to be stable, and be the cation . Silver has only one electron too, so the cation will be
.
To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:
Sodium carbonate: Na2CO3
Silver nitrate: AgNO3
Silver carbonate: Ag2CO3
Sodium nitrate: NaNO3
The balanced reaction will be:
Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3
Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)
The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:
Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol
So the molar mass of the compounds must be:
Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)
AgNO3 = 170 g/mol (108 + 14 + 3x16)
Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)
NaNO3 = 85 g/mol
We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:
1 mol of Na2CO3 ---------- 2 mol of AgNO3
106 g ------------------------------ 2x170 = 340 g
x ------------------------------------ 5.14 g
By a simple direct three rule:
340x = 544.84
x = 1.6 g of Na2CO3
That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.
1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:
m = 3.6 - 1. 6 = 2.0 g
2) All the AgNO3 reacted, so there isn't a mass present after the reaction.
m = 0 g
3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3
2 moles of AgNO3 ------------- 1 mol of Ag2CO3
2x170 g ------------------------------- 276 g
5.14 g --------------------------------- x
By a simple direct three rule:
340x = 1418.64
x = 4.17 g of Ag2CO3
4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3
2 moles of AgNO3 ----------------------- 2 moles of NaNO3
2x170 g ---------------------------------------- 2x85 g
5.14 g ------------------------------------------- x
By a simple direct three rule:
340x = 873.8
x = 2.57 g
Answer:
The solution will turn red.
Explanation:
HC₁₄H₁₄SO₃ + H₂O ⇌ HC₁₄H₁₄SO₃⁻ +H₃O⁺
(red) (yellow)
Methyl orange is a weak acid in which the ionized and unionized forms are distinct colours and are in equilibrium with each other,
At about pH 3.4, the two the forms are present in equal amounts, and the indicator colour is orange.
If you add more acid, you are disturbing the equilibrium.
According to Le Châtelier's Principle, when you apply a stress to a system at equilibrium, it will respond in such a way as to relieve the stress.
The system will try to get rid of the added acid, so the position of equilibrium will move to the left.
More of the unionized molecules will form, so the solution will turn red.
