The statements in accordance with the law of conservation of charge are:
A. The total charge of the reactants and products must be equal
B. The net charge of an isolated system remains constant
Both of these statements follow the law of conservation of charge which states that charge may neither be created nor destroyed, due to which the total charge in an isolated system (one in which charge can not move in or out of) remains constant.
1) Find the number of moles that the final solution must contain
M = n / liters of solution => n = M*liters of solution
n = 1.5 mol/liter * 25.0 liter = 37.5 moles
2) Find how many liters of the stock solution contain 37.5 moles of HCL
M = n / liters of sulution => liters of solution = n / M = 37.5 mol / 18.5 mol/liter
liter of solution = 2.03 liter
Answer: 2.03 liter
Answer:
2.40 × 10²³ Atoms of Hydrogen
Solution:
The number of hydrogen atoms in C₃H₈O₃ are calculated as;
As, there are eight hydrogen atoms per molecule of C₃H₈O₃, therefore the number of hydrogen atoms per mole of C₃H₈O₃ are calculated as;
1 mole C₃H₈O₃ contains = 6.022 × 10²³ Molecules of C₃H₈O₃
So,
0.05 mole C₃H₈O₃ will contain = X Molecules of C₃H₈O₃
Solving for X,
X = (0.05 × 6.022 × 10²³) ÷ 1
X = 3.011 × 10²² Molecules of C₃H₈O₃
As,
1 Molecule of C₃H₈O₃ contains = 8 Atoms of Hydrogen
So,
3.011 × 10²² Molecules of C₃H₈O₃ will contain = X Atoms of Hydrogen
Solving for X,
X = (3.011 × 10²² × 8) ÷ 1
X = 2.40 × 10²³ Atoms of Hydrogen
Answer: The student’s values are accurate as well as precise.
Explanation:
Precision refers to the closeness of two or more measurements to each other.
For Example: If you weigh a given substance three times and you get same value each time. Then the measurement is very precise.
Accuracy refers to the closeness of a measured value to a standard or known value.
For Example: If the mass of a substance is 50 kg and one person weighed 49 kg and another person weighed 48 kg. Then, the weight measured by first person is more accurate.
Given: Mass = 5.000 g
Mass weighed by A has values 4.891 g , 4.901 g and 4.890. Thus the average value is 
Thus as the measured value is close to the true value, the student’s values are accurate and as the values are close to each other, the measurement is precise.
