Answer:
0.978 M
Explanation:
Given data
- Mass of luminol (solute): 13.0 g
- Volume of the solution = volume of water: 75.0 mL = 0.0750 L
We can find the molarity of the stock solution of luminol using the following expression.
M = mass of solute / molar mass of solute × liters of solution
M = 13.0 g / 177.16 g/mol × 0.0750 L
M = 0.978 M
Answer:
1.) 5 or D
2.) 2 or B
3.) 4 or D
4.) 1 or A
Explanation:
I got them correct on the quiz on edge
Answer:
2.038 seconds.
Explanation:
So, in the question above we are given the following parameters in order to solve this question. We are given a rate constant of 0.500 s^-, initial concentration= 0.860 M and final concentration= 0.310 M,the time,t =??.
Assuming that the equation for the first order of reaction is given below,that is;
A ---------------------------------> products.
Recall the formula below;
B= B° e^-kt.
Therefore, e^-kt = B/B°.
-kt = ln B/B°.
kt= ln B°/B.
Where B° and B are the amount of the initial concentration and the amount of the concentration remaining, k is the rate constant and t = time taken for the concentration to decrease.
So, we have; time taken,t = ln( 0.860/.310)/0.500.
==> ln 2.77/0.500.
==> time taken,t =2.038 seconds.
The number of moles of ethanol the chemist will use in the experiment involving 30g of ethanol is 0.65moles.
<h3>How to calculate number of moles?</h3>
The number of moles of a substance can be calculated by dividing the mass of the substance by its molar mass. That is;
no. of moles = mass ÷ molar mass
According to this question, a chemist will use a sample of 30 g of ethanol (CH3CH2OH) in an experiment. The number of moles can be calculated as follows:
Molar mass of ethanol = 12(2) + 1(5) + 17 = 46g/mol
no of moles = 30g ÷ 46g/mol
no. of moles = 0.65moles
Therefore, the number of moles of ethanol the chemist will use in the experiment involving 30g of ethanol is 0.65moles.
Learn more about moles at: brainly.com/question/1458253
Answer:
Of lower concentration or less concentrated
Explanation:
Osmosis is the movement of solvent from a region of lower concentration of solute to a region of higher concentration of solute through a semipermeable membrane in order to equalize the concentration of the solutions on both sides.
Since the membrane of the bag is semipermeable, then the fact that the bag in the beaker decreased in size, lost volume, and became flaccid indicates that the solution in the bag is of lower solute concentration than the solution in the beaker hence the movement of water molecules into the beaker by osmosis.
