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Alik [6]
2 years ago
14

A beaker contains 0.125 L of a 3.00 M solution. If the volume goes up to 0.325 L, what is the new molarity?

Chemistry
1 answer:
igor_vitrenko [27]2 years ago
8 0

Answer:

1.15 M

Explanation:

Step 1: Given data

  • Initial volume (V₁): 0.125 L
  • Initial concentration (C₁): 3.00 M
  • Final volume (V₂): 0.325 L
  • Final concentration (C₂): ?

Step 2: Calculate the final concentration of the solution

We want to prepare a dilute solution from a concentrated one by adding water. We can calculate the concentration of the dilute solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 3.00 M × 0.125 L/0.325 L = 1.15 M

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Answer:

This law states that, despite chemical reactions or physical transformations, mass is conserved — that is, it cannot be created or destroyed — within an isolated system. In other words, in a chemical reaction, the mass of the products will always be equal to the mass of the reactants.

Explanation:

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Electrophilic addition of bromine, Br2, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermedia
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Answer:

See explanation and image attached for details

Explanation:

The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.

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3 years ago
The following reaction has an equilibrium constant (Keq) of 1 under standard conditions and can be catalysed by the enzyme aspar
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Answer:

  1. What is the AGⓇ of this reaction? 0.
  2. Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
  3. What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
  4. If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.

Explanation:

1. To calculate the delta G of a reaction given the K, we use the following equation:

ΔG°= -RT ln K.

Which gives us 0 when K is 1.

2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.

3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.

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2 years ago
Janice bought 40 shares of stock at $31.82 per share. She received dividends of $1.11 per share for 1 year. (Do not use commas,
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Answer:

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Explanation:

The <em>purchase price</em> is what Janice invested for every share.

Since the stock was priced at $31.82 per share and she received a $1.11 dividend per share, her investment was:

  • $31.82 - $1.11 = $30.71 per share ← answer

This price is the cost for Janice, over which she shall calculate their returns (gains or losses) on the future, when she sells the shares, for instance.

The total investment of Janice was the number of shares multipled by the purchase price:

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Carson’s class visited a hydroelectric plant on a field trip during their study of energy. The process interested him, and he ha
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Answer:

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Explanation:

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