Answer:
Answers provided for the reactions with explanations in parenthesis
Explanation:
We need to find the approx. equivalent pH so we can choose an indicator.
CH3COONa +strong acid HCl - methyl orange (pH will be about 4.5)
Na2CO3 base ---> strong acid HCl- phenolphthalein (pH will be 7-8) , and then methyl orange (since carbonic acid has 2 equivalence points, it needs two indicators)
Na2C2O4 HCl- methyl orange, and then methyl orange (since oxalic acid has 2 equivalence points, it needs two indicators)
NH4Cl NaOH- Phenolphthalein, since pH = 9.75 approximately
NaOH HCl-Bromothymol blue --> since pH = 7 best for a neutral indicator
The percentage of the sulfur (S) in the compound CuSO₄ is 20.1 %.
<h3>What is the mass percentage?</h3>
The percentage of an element in a compound can be determined as the number of parts by mass of that element present in 100 parts by mass of the given compound.
First, calculate the molecular mass of the given compound by the addition of the atomic masses of all the present elements in the molecular formula. Then, the percentage of the elements can be determined by dividing the total mass of the element by the molar mass of the compound multiplied by 100.
Given, the atomic mass of copper, sulfur, and oxygen is 63.55 g, 32.07 g, and 16.0g respectively.
The molecular mass of CuSO₄ = 63.55 + 32.07 + 4(16.0) = 159.62 g
The mass percentage of the sulphur = (32.07/159.62) × 100 = 20.1 %
Therefore, the mass percentage of the sulfur is equal to 20.1 %.
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Friction is helpful when riding on dirt or cement but it isn't when in rain or mud.
Answer:
The coefficient of O2 is 11
Explanation:
Step 1:
The equation for the reaction:
FeS2 + O2 → SO2 + Fe2O3
Step 2:
Balancing the equation. The equation can be balance as follow:
FeS2 + O2 → SO2 + Fe2O3
There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:
2FeS2 + O2 → SO2 + Fe2O3
There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:
2FeS2 + O2 → 4SO2 + Fe2O3
Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:
2FeS2 + 11/2O2 → 4SO2 + Fe2O3
Multiply through by 2 to clear the fraction as shown below:
4FeS2 + 11O2 → 8SO2 + 2Fe2O3
Now the equation is balanced.
The coefficient of O2 is 11