Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:
1. Weigh all the equipment and materials required in the experiment before the experiment.
2. Avoid spillage and evaporation during the experiment.
3. Weigh all the equipment and materials after the experiment.
If the mass is conserved then weight from step 1 is equal to weight from step 3.
Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk
Energy content of 3 oz skinless chicken is = 110 kcal
Energy content of 3 oz broccoli = 30 kcal
Energy content of 1 medium apple = 60 kcal
Energy content of 1 cup non-fat milk = 90 kcal
So the kilocalories of energy patient obtained from lunch
= 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal
Answer:
D. 15g
Explanation:
The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.
In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.
Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
Answer: The density of the object will be 
Explanation:
Density is defined as the mass contained per unit volume.
Given : Mass of object = 19.6 grams
Volume of object= 
Putting in the values we get:
Thus density of the object will be
