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Licemer1 [7]
3 years ago
5

A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solutio

n is 0.046 M KOH, but has forgotten to record the volume of the original sample. The concentration of the original solution is 2.09 M. What was the volume of the original sample?
Chemistry
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

2.2mL

Explanation:

First, let us analyse what was given from the question:

C1 = 2.09M

V1 =?

C2 = 0.046M

V2 =100mL

Using dilution formula (C1V1 = C2V2), we can calculate the volume of the original solution as follows:

C1V1 = C2V2

2.09 x V1 = 0.046 x 100

Divide both side by the coefficient of V1 ie 2.09, we have:

V1 = (0.046 x 100) / 2.09

V1 = 2.2mL

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Solve pls brainliest
sammy [17]

Answer:

Even though the two substances possess many similarities, they have some unique properties. In turn, since they have the same properties, if they were the same substance, it would make matters worse, if the same chemical was in two different places, there would not be a difference between them since they are the same, just as it is with are two different chemicals would have differing properties since they are two properties would vary from one another since they are 2 totally different things!

4 0
2 years ago
What is the enthalpy of combustion when 1 mol C6H6(g) completely reacts with oxygen? 2C6H6(g) 15O2(g) mc001-1. Jpg 12CO2(g) 6H2O
Andrei [34K]

The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .

The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.

The thermochemical equation for the combustion of benzene is;

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol

We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.

Learn more: brainly.com/question/13164491

8 0
2 years ago
why a gas, at constant pressure, increases greatly in volume when it is gently heated but a solid does not
11Alexandr11 [23.1K]

Explanation:

This is because gas particles are free to move as they are not held in place by strong molecular forces while particles in a solid are

8 0
3 years ago
Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2
Gekata [30.6K]

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide (CO_2) compound

The combination of 2 blue spheres will represent nitrogen molecule (N_2)

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide (NO) compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide (NO) compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = \frac{2}{2}\times 4=4mol of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0
3 years ago
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
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