Explanation:
The given precipitation reaction will be as follows.

Here, AgCl is the precipitate which is formed.
It is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.
Molarity = 
0.269 M = 
no. of moles = 0.306 mol
As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.
No. of moles = 
0.307 mol = 
mass = 43.99 g
Thus, we can conclude that mass of precipitate produced is 43.99 g.
Answer:
the Law of multiple proportions
Explanation:
The law of multiple proportions states that, if two elements A and B combine to form more than one chemical compound, then the various masses of one of the elements A, which combines with a fixed mass of the element B are in simple multiple ratio.
This is demonstrated in the formation of nitrogen compounds such as NO and N2O when nitrogen combines with oxygen. This ratio is always constant.
Answer:
C. A Spring Scale
Explanation:
Using process of elimination, we can quickly decide that a stopwatch and a ruler will not be useful in measuring the force. This leaves us with either the spring scale or a balance scale. A balance scale is used to compare two weights, so this is eliminated. That leaves us with a spring scale. This is because we can attached the spring scale to the car and when we let it go, we can record the force. I hope this helps!
Answer:
the answer is atoms and molecules
Answer:
A. 1.54 mole.
B. 55.39g of carbon
Explanation:
A. Determination of the number of mole in 67.7g of C3H8.
Mass of C3H8 = 67.7g
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Number of mole of C3H8 =..?
Number of mole = Mass/Molar Mass
Number of mole of C3H8 = 67.7/44
Number of mole of C3H8 = 1.54 mole
B. Determination of the mass of carbon in the compound.
This is illustrated below:
The mass of C in compound can be obtained as follow:
=> 3C/C3H8 x 67. 7
=> 3x12 / 44 x 67.7
=> 36/44 x 67.7
=> 55.39g
Therefore, 55.39g of carbon is present in the compound.