The fridge part can, just not the freezer, I think.
Answer:
459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution
Explanation:
Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by the following expression:

Molarity is expressed in units 
In this case:
- Molarity: 1.56 M= 1.56

- Number of moles of calcium chlorine= ?
- Volume= 2.657 liters
Replacing:

Solving:
Number of moles of calcium chlorine= 1.56 M* 2.657 liters
Number of moles of calcium chlorine= 4.14 moles
In other side, you know:
- Ca: 40 g/mole
- Cl: 35.45 g/mole
Then the molar mass of the calcium chloride CaCl₂ is:
CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole
Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

mass= 459.126 g
<u><em>459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution</em></u>
Answer:
0.4
Explanation:
Given data:
Number of moles of SrCl₂ consumed = ?
Mass of ZnCl₂ produced = 54 g
Solution:
Chemical equation:
ZnSO₄ + SrCl₂ → SrSO₄ + ZnCl₂
Number of moles of ZnCl₂:
Number of moles = mass/ molar mass
Number of moles = 54 g/136.3 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of ZnCl₂ with SrCl₂ from balance chemical equation.
ZnCl₂ : SrCl₂
1 : 1
0.4 : 0.4
Thus when 54 g of ZnCl₂ produced 0.4 moles of SrCl₂ react.
Phenolphthalein turn pink in basic solutions.it turn colourless in acid solution.
Hence options(1)is correct
We can consider a chemical to be safe if it does not contain any harmful substances such as nitrogen, harmful acids, or even excessive heat. Other chemicals such as the chemicals used in food products like, for example, citric acid, is not so harmful.
Hope this helps! :D
Any questions? Just let me know! I'd be happy to help any way possible.