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2 years ago

To make a 6.50M solution, how many moles of solute will be needed if 12.0 liters of solution are required

Chemistry

1 answer:

avanturin [10]2 years ago

8 0

Answer:

78 moles of the solute

Explanation:

From the question;

Molarity of the solution is 6.50 M
Volume of the solution is 12.0 L

We want to determine the number of moles needed

We need to know that;

Molarity = Number of moles ÷ Volume

Therefore;

Number of moles = Molarity × Volume

Hence;

Number of moles = 6.50 M × 12.0 L

= 78 moles

Thus, the moles of the solute needed is 78 moles

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rodikova [14]

Answer:

M=Molarity

V=volume

change millilitre to litre

so:15ml=0.015 L

:38.5ml=0.0385 L

M1V1=M1V1

M1* 0.015 L= 0.15M*0.0385 L

M1=0.00385

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3 years ago

1. Baking soda and vinegar combined *

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Answer:

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Explanation:

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2 years ago

In a redux reaction why does an element oxidation number decreases

Olin [163]

Answer:

Decreases

Explanation:

An element's oxidation number decreases because it gained electrons.

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2 years ago

"Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units

olasank [31]

Answer:

Part a: Units of k is $M^{-2}s^{-1}$ where reaction is first order in A and second order in B

Part b: Units of k is $M^{-1}s^{-1}$ where reaction is first order in A and second order overall.

Part c: Units of k is $M^{-1}s^{-1}$ where reaction is independent of the concentration of A and second order overall.

Part d: Units of k is $M^{-3}s^{-1}$ where reaction reaction is second order in both A and B.

Explanation:

As the reaction is given as

$A+B \rightarrow C$

where as the rate is given as

$r=k[A]^x[B]^y$

where x is the order wrt A and y is the order wrt B.

Part a:

x=1 and y=2 now the reaction rate equation is given as

$r=k[A]^1[B]^2$

Now the units are given as

$r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k$

The units of k is $M^{-2}s^{-1}$

Part b:

x=1 and o=2

x+y=o

1+y=2

y=2-1

y=1

Now the reaction rate equation is given as

$r=k[A]^1[B]^1$

Now the units are given as

$r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part c:

x=0 and o=2

x+y=o

0+y=2

y=2

Now the reaction rate equation is given as

$r=k[A]^0[B]^2$

Now the units are given as

$r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part d:

x=2 and y=2

Now the reaction rate equation is given as

$r=k[A]^2[B]^2$

Now the units are given as

$r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k$

The units of k is $M^{-3}s^{-1}$

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3 years ago

What two parts of an amino acid are involved in a peptide bond?

Lemur [1.5K]

Answer:

The answer to your question is below.

Explanation:

Amino acids are composed by one amino group, one carboxyl group and one chain.

The parts of the amino acid that are involved in a peptide group are the amino group (- NH₂) and the carboxyl group (-COOH).

6 0

3 years ago