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2 years ago

Which of following is true of liquids?

Chemistry

2 answers:

Mashutka [201]2 years ago

7 0

Answer:

The particles move at a high rate of speed.

The particles have a medium amount of kinetic energy and slide past one another.

Explanation:

The options that are true about liquids are A. The particles move at a high rate of speed and B. The particles have a medium amount of kinetic energy and slide past one another, because particles in a liquids are slippery, so they slide past one another. This is how liquids move around. The particles help it move around.

zvonat [6]2 years ago

6 0

Answer:

The particles have a medium amount of kinetic energy and slide past one another

Explanation:

If particles move at a high rate of speed its gas, and if they vibrate in place and they have a fixed volume and shape its probably a solid. So a medium amount of kinetic enery is correct.

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Answer: because ch4 is not considered a acid they said it is too weak

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3 years ago

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3 years ago

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Alex Ar [27]

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3 years ago

Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and

notka56 [123]

Answer:

$pH=4.56$

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

$n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol$

And the concentrations are:

$[acid]=0.000855mol/0.025L=0.0342M$

$[base]=0.000781mol/0.025L=0.0312M$

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

$pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56$

Best regards!

6 0

2 years ago

At 311 K, this reaction has a K c value of 0.0111 . X ( g ) + 2 Y ( g ) − ⇀ ↽ − 2 Z ( g ) Calculate K p at 311 K. Note that the

Aleks [24]

Answer:

$K_{p}=4.35\times 10^{-4}$

Explanation:

We know, $K_{p}=K_{c}(RT)^{\Delta n}$

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and $\Delta n$ is difference in sum of stoichiometric coefficient of products and reactants

Here $\Delta n=(2)-(2+1)=-1$ and T = 311 K

So, $K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}$

Hence value of equilibrium constant in terms of partial pressure $(K_{p})$ is $4.35\times 10^{-4}$

4 0

2 years ago