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3 years ago

50POINTS!

Chemistry

2 answers:

stiks02 [169]3 years ago

8 0

Answer:

Hi. I'll explain

Explanation:

$11.2mol \times \frac{22.4liter}{1mol} = 250.88$

If you round it it will be 251

D is the answer (251)

Lelu [443]3 years ago

7 0

The answer is D 251L hope you get a good grade

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Please help!!!!

Tomtit [17]

Answer 1) Option D : Discrete

Explanation : As in the example the scientist studies about the effects of growing human populations on the biodiversity which is found in a region, and where each region was selected had a different population density of humans from 1 to 10 million per 10 square miles. Then in each region the number of different species that can be found was recorded. So here the numerical data was collected for different regions. So, we can conclude it as discrete because when the variable takes on a countable number of values it is called as discrete.

Answer 2) Option D : The entertainment industry

Explanation : When people had enough time and money with them there was no need of creating the communications industry as it was a need not a luxurious thing. With the creation of Internet industry it is somewhat irrelevant. They had enough of time and money so building a labor industry seems to be a lame choice. So, the option of entertainment industry suits the best.

6 0

3 years ago

Read 2 more answers

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

3 years ago

If a light bulb is missing or Broken in a parallel circuit,will the other bulb light

umka2103 [35]

Yes, if it’s a parallel circuit the wires are two different wires so it will light because that bulb isn’t connected to the one that went out

5 0

3 years ago

Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expres

Marizza181 [45]

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

$Ksp=[Ag^+][Cl^-]$

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

I - 0 0

C - +x +x

E - x x

Which leads to the following modified equilibrium expression:

$Ksp=x^2$

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

6 0

3 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

5 0

3 years ago