Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
Answer:
Following are the answer to this question:
Explanation:
In the given question information is missing, that is equation which can be defined as follows:

- Growing temperatures may change its connection to just the way which consumes thermal energy in accordance with Le chatelier concepts Potential connection is endothermic. Answer: shifts to the right
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Kc are described as a related to the concentration by the intensity of both the reaction for each phrase which reaches a power equal towards its stoichiometric equation coefficient Kc = \frac{product}{reactant}
It increases [product] but reduces [reactant] Therefore, Kc increases
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.
Answer:
4 g after 58.2 years
0.0156 After 291 years
Explanation:
Given data:
Half-life of strontium-90 = 29.1 years
Initially present: 16g
mass present after 58.2 years =?
Mass present after 291 years =?
Solution:
Formula:
how much mass remains =1/ 2n (original mass) ……… (1)
Where “n” is the number of half lives
to find n
For 58.2 years
n = 58.2 years /29.1 years
n= 2
or 291 years
n = 291 years /29.1 years
n= 10
Put values in equation (1)
Mass after 58.2 years
mass remains =1/ 22 (16g)
mass remains =1/ 4 (16g)
mass remains = 4g
Mass after 58.2 years
mass remains =1/ 210 (16g)
mass remains =1/ 1024 (16g)
mass remains = 0.0156g
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?



volume of water added = final volume -initial volume
= 120-10
=110ml