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Lera25 [3.4K]
2 years ago
12

A group of students want to see how temperature affects the time spilled water to dry up what will be independent and dependent

variable
Chemistry
1 answer:
erastova [34]2 years ago
4 0
Independent is the temperature dependent is the time
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Question 23
Schach [20]

Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.

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Calculate the pH of the resulting solution if 26.0 mL of 0.260 M HCl(aq) is added to:
enyata [817]
So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
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Write the molecular formula for the following compound.<br>​
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Answer : The molecular formula of the compound is, C_8H_9O_3N

Explanation :

Molecular formula : It is the representation of substance by the symbols and it denotes the number of atoms of each element present in the compound.

Now count the number of carbon, hydrogen, nitrogen and oxygen atoms present in the given compound.

As we see that in the given compound, there are 8 atoms of carbon element, 3 atoms of oxygen element, 1 atom of nitrogen element, 9 atoms of hydrogen element.

Thus, the molecular formula of the compound will be C_8H_9O_3N

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