1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
Molar concentration of the weak acid solution is 0.0932
Explanation:
Using the formula: 
Where Ca = molarity of acid
Cb = molarity of base = 0.0981 M
Va = volume of acid = 25.0 mL
Vb = volume of base = 23.74 mL
na = mole of acid
nb = mole of base
Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1
Therefore, 
Ca = 0.0981 x 23.74/25.0
= 0.093155 M
To 4 significant figure = 0.0932 M
Answer:
When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.
This decreases the blood glucose level in the level. This is how the negative feedback works in the body if the level of glucose increases.
Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.
We first calculate for the number of moles of NaOH by dividing the given mass by the molar mass of NaOH which is equal to 40 g/mol. Solving,
moles of NaOH = (68.4 g/ 40 g/mol) = 1.71 moles NaOH
Then, we divide the calculate number of moles by the volume in liters.
molarity = (1.71 moles NaOH / 0.875 L solution)
molarity = 1.95 M
Answer:
Option B
Explanation:
We will check the solubility graph for potassium nitrate, KNO
3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.
Hence, option B is correct