Answer:
1.99 x 10²³ formula units
Explanation:
Given parameters:
Mass of AgF = 42.15g
Unknown:
The amount of formula units
Solution:
To solve this problem, we set out by find the number of moles in this compound from the given mass.
Number of moles = 
molar mass of AgF = 107.9 + 19 = 126.9g/mol
Number of moles = 
= 0.33 moles
1 mole of a substance = 6.02 x 10²³ formula units
0.33 moles of AgF = 0.33 x 6.02 x 10²³ = 1.99 x 10²³ formula units
Explanation:
To remove the salt from the oil, I will add water to dissolve the salt from it.
Oil is an organic molecule that is non-polar
Salt is polar ionic compound
Salt will not dissolve in the oil.
- Take the mixture.
- Add water to it.
- Water and oil are immiscible
- Shake the new heterogeneous mixture vigorously.
- leave to settle.
- Oil will come on top of the water.
- You can skim off the oil layer on top.
- Then heat the water and salt solution.
- This leaves the oil behind.
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Answer:
The answer to your question is: number 1
Explanation:
Third law of Newton: says that for every action ,there is an equal and opposite reaction.
So, if the child is pushing an object to the right, it will recipe the same amount of force that he is exerting to the object but in opposite direction.
Number 2 and 3 are incorrect because, because the third law of Newton says "an equal and opposite reaction", not slightly more or less.
Number 4 is wrong, it is not in agreement with Newton's third law of motion.
The reaction will produce 12.1 g Ag₂S.
<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S
<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)
× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
