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Lemur [1.5K]
2 years ago
12

If an acid is added to different samples of buffered water, the sample with the

Chemistry
2 answers:
Eddi Din [679]2 years ago
5 0

Answer:

react with acid that is added and make a base

Explanation:

Buffer solutions are made when a solution of weak acid is mixed with its conjugate base.

Here the buffer solution will react with the strong acid .

Helen [10]2 years ago
3 0

Answer:

<h3><u>A). react with acid that is added and make a base.</u></h3>

explanation:

<em>Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added.</em>

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A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.601 mol of a gaseous product. If the cyli
netineya [11]

Answer:

4.33 L

Explanation:

Assuming ideal behaviour and that all 0.300 moles of gas reacted, we can solve this problem using Avogadro's law, which states that at constant temperature and pressure:

  • V₁n₂ = V₂n₁

Where in this case:

  • V₁ = 2.16 L
  • n₂ = 0.601 mol
  • V₂ = ?
  • n₁ = 0.300 mol

We <u>input the given data</u>:

  • 2.16 L * 0.601 mol = V₂ * 0.300 mol

And <u>solve for V₂</u>:

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3 years ago
What chemical property must argon have to make it a suitable choice for light bulbs?
Sedbober [7]
Argon is a suitable choice for light bulbs because it is inert. Compared to a reactive gas like oxygen, the metal filimant would burn up in a reactive environment, which is why a noble gas is used.
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3 years ago
What is potential energy
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basically just energy held by an object because of its relation to other objects

Explanation:

7 0
3 years ago
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol
Kryger [21]

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant K_c=  \dfrac{[CO][H_2]}{[H_2O]}

The equilibrium constant  K_c=  \dfrac{(0.17 )(0.17)}{0.74}

The equilibrium constant K_c=  0.03905

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \  \ \ \ \ \to0.17

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

K_c=  \dfrac{[CO][H_2]}{[H_2O]}

0.03905 =  \dfrac{[0.17+x][x]}{[0.91 -x]}

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265  or   x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

3 0
4 years ago
What steps should you take to respond to an accident?
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2 years ago
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