It’s called Diethyl ether so I think the answer is D
Answer:
See Explanation
Explanation:
The question is incomplete; as the mixtures are not given.
However, I'll give a general explanation on how to go about it and I'll also give an example.
The percentage of a component in a mixture is calculated as:
Where
E = Amount of element/component
T = Amount of all elements/components
Take for instance:
In 
The amount of all elements is: (i.e formula mass of )
The amount of calcium is: (i.e formula mass of calcium)
So, the percentage component of calcium is:
The amount of hydrogen is:
So, the percentage component of hydrogen is:
Similarly, for oxygen:
The amount of oxygen is:
So, the percentage component of oxygen is:
Answer:
curium
−
243
,
252
/
99
Es,
251
/
98
Cf,
214
/
82
Pb
Explanation: Im not very good with this but here ya go!
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
Answer:
0,508g of H₂O₂
Explanation:
For the reaction:
2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) → 3O₂(g) + 2MnSO₄(aq) + K₂SO₄(aq) + 4H₂O(l)
2 moles of KMnO₄ react with 1 mol of H₂O₂.
In the titration, moles of KMnO₄ required were:
1,68M×0,0178L = 0,0299 moles of KMnO₄. Moles of H₂O₂ are:
0,0299 moles of KMnO₄×
= 0,01495 moles of H₂O₂. As molar mass of H₂O₂ is 34,01g/mol, mass of H₂O₂ was dissolved is:
0,01495 moles of H₂O₂×
= <em>0,508g of H₂O₂</em>
