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kirill115 [55]
3 years ago
11

Juan Carlos placed 35 grams of ke into a dry, 200-gram container. The top of the container was attached tightly. When the ice wa

s completely melted, the student
A.35
B.165
C.200
D.235

20 POINTS!!
Chemistry
1 answer:
lianna [129]3 years ago
8 0
Correct answer is D
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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
koban [17]

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 CO3^{-2}, and the ion nitrate is NO3^{-}.

Sodium is in group 1, so it must lose one electron to be stable, and be the cation Na^{+}. Silver has only one electron too, so the cation will be Ag^{+}.

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

8 0
3 years ago
Answer that pls will give brainlist
ELEN [110]

Answer:

1. went on the first manned launch during the Apollo mission

2. responsible for all spacecraft systems

3. 4500 hour

4. published All-American Boys

5. Joined back in space organization as a consultant to inspire next generation for mars

4 0
3 years ago
Washing machines use a large amount of water. A student suggested that old pairs of stained jeans which have to be washed more f
sukhopar [10]

Answer:

C |||| It is not practical because it takes a huge amount of water to produce a new pair of jeans

Explanation:

If you're doing flvs then it's C.

5 0
3 years ago
Read 2 more answers
Describe the capillary action in acetone compared to water
lakkis [162]
Capillary action is defined as the ability of a liquid to go up a narrow space without the help or opposition of external forces. One of the most important factors affecting capillary action is the intermolecular forces within a substance. The higher the IMF, the greater the capillary action. The H-bonding in water gives it greater IMF than acetone, so water has greater capillary action.
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3 years ago
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) palmitic acid is a 16 carbon acid. in a balanced equation, the products of the saponification of glyceryl tripalmitate (tripal
lesya [120]

Answer: The products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

Explanation:

A chemical reaction in which triglycerides react with sodium hydroxide and leads to the formation of one molecule of glycerol and three molecules of a salt of fatty acid is known as saponification.

For example, when tripalmitin reacts with sodium hydroxide then it leads to the formation of one molecule of glycerol and three molecules of sodium salt of palmitic acid.

The reaction equation is as follows.

   C_{51}H_{98}O_{6} + NaOH \rightarrow C_{3}H_{8}O_{3} + 3C_{16}H_{31}O_{2}Na^{+}

Thus, we can conclude that the products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

8 0
3 years ago
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