The percent composition of each element can be calculated as follows:
% composition = (mass of element / total mass) * 100
The total mass of the quarter is given to be 5.670 grams
Mass of Cu = 5.198 grams
Mass of Ni = 0.472 grams
Substitute in the above equation to get the mass percentage of each element as follows:
% of Cu = (5.198/5.670) * 100 = 91.675%
% of Ni = (0.472/5.670) * 100 = 8.325%
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
The mass of coke needed to react completely with 1.0 ton of copper(II) oxide is 0.794 Ton.
Explanation:

1 Ton = 907185 grams
Mass of copper oxide = 1.0 Ton = 907185 grams
Moles of copper oxide =
According to reaction, 2 moles of copper oxide reacts with 1 mole of carbon.
Then 11403.95 moles of copper oxide will react with:
of carbon
Mass of 5,701.98 moles of carbon:

Mass of coke = x
Mass of carbon = 68,423.75 g
Percentage of carbon in coke = 95%


The mass of coke needed to react completely with 1.0 ton of copper(II) oxide is 0.794 Ton.
Answer:
False
Explanation:
Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.
<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>
Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.
Answer:
1422mg of acetaminophen
Explanation:
The maximum dose of acetaminophen is 15.0 mg of acetaminophen per kg of person.
To know the maximum single dosage of the person we need to convert the 209lb to kg (Using 1kg = 2.2046lb):
209lb * (1kg / 2.2046lb) = 94.8
The person weighs 94.8kg and the maximum single dosage for the person is:
94.8kg * (15.0mg acetaminophen / kg) =
1422mg of acetaminophen