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kvasek [131]
3 years ago
15

Part a: What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl?

Chemistry
1 answer:
Tasya [4]3 years ago
6 0
Hey! Let me help you!

C)

<span>What additional volume of 10.0 M HCl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B? In other words, how much more of this HCl solution is required to consume the remaining Tris in the buffer?

let x = how much more solution needed

x = 1.8 mL <=========

Your answer is 1.8 mL as it is needed for more of this HCI solution is requires to consume the remaining Tris in the buffer!

Have an awesome day! :D
</span>
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When a base is dissolved in water, what is released?<br> A. H+<br> B. OH−<br> C. CO2
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Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces
Alchen [17]

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

<u>Answer:</u> The pressure equilibrium constant for the reaction is 32908.46

<u>Explanation:</u>

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

                    4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)

<u>Initial:</u>               2.2          2.4

<u>At eqllm:</u>        2.2-4x      2.4-3x         2x        6x

Evaluating the value of 'x':  

\Rightarrow 2x=0.99\\\\x=0.495

So, equilibrium partial pressure of ammonia = (2.2 - 4x) = [2.2 - 4(0.495)] = 0.22 atm

Equilibrium partial pressure of oxygen gas = (2.4 - 3x) = [2.4 - 3(0.495)] = 0.915 atm

Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm

The expression of K_p for above equation follows:

K_p=\frac{(p_{N_2})^2\times (p_{H_2O})^6}{(p_{NH_3})^4\times (p_{O_2})^3}  

Putting values in above equation, we get:

K_p=\frac{(0.99)^2\times (1.98)^6}{(0.22)^4\times (0.915)^3}\\\\K_p=32908.46

Hence, the pressure equilibrium constant for the reaction is 32908.46

5 0
3 years ago
A piece of aluminum has a volume of 12 cm3and a mass of 32 g. What is its density? Do not forget the units.
Contact [7]

Answer:

<h3>The answer is 2.67 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

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mass of metal = 32 g

volume = 12 cm³

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Jobisdone [24]
The answer is 6
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