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11Alexandr11 [23.1K]

2 years ago

10

A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158

K. Which of the following are true? Choose all that apply
a. The liquid initially present will solidify.
b. The final state of the substance is a gas.
c. The sample is initially a solid.
d. The sample is initially a liquid. One or more phase changes will occur.

1 answer:

Zinaida [17]2 years ago

4 0

Answer:

b. The final state of the substance is a gas.

d. The sample is initially a liquid. One or more phase changes will occur.

Explanation:

Methane has the following properties:

Normal melting point: 90.7 K
Normal boiling point: 111.65 K

*"Normal" refers to normal pressure (1 atm).

According to this, we can affirm:

Below 90.7 K, methane is solid.
Between 90.7 K and 111.65 K, methane is liquid.
Above 111.65 K, methane is gas.

<em>A sample of methane at a pressure of 1.00 atm and a temperature of 93.1 K is heated at constant pressure to a temperature of 158 K. Which of the following are true? Choose all that apply.</em>

<em>a. The liquid initially present will solidify.</em> FALSE. The liquid will vaporize.

<em>b. The final state of the substance is a gas.</em> TRUE.

<em>c. The sample is initially a solid.</em> FALSE. The sample is initially a liquid.

<em>d. The sample is initially a liquid. One or more phase changes will occur. </em>TRUE.

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Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

<u>(e) a $1\times 10^{-4}M$ solution of $HNO_2$ </u>

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

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Electromagnetic waves heat earths surface through convection?

larisa86 [58]

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3 years ago

Why is it important for scientists to review and repeat the work of other scientists?

Veronika [31]

Scientists repeat experiments for reliability. Experiments have to be repeated, since performing an experiment only once, does not prove a scientists theory on the experiment, which they develop by performing the experiment. And then performing the experiment again or even many times to prove or disapprove their theories. Btw, before an experiment begins, the scientist will make a hypothesis of what they believe will happen. If proven correctly, they would then use those results they record throughout the experiment, from beginning to end to prove whether or not their hypothesis are correct or incorrect. Click to let others know, how helpful is it

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So that they can prove the other scientists views wrong and develop a new theory in which there is more sense and credibility.

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BASICALLY, I THINK THE ANSWER IS 1 OR 2 OR

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3 years ago

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The equilibrium constant, Kc, for the following reaction is 3.61×10-4 at 426 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar

iogann1982 [59]

Answer:

the concentration of PCl5 in the equilibrium mixture = 296.20M

Explanation:

The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.

The detaied steps is as shown in the attached file.

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IUPAC NAME FOR:<br>CH2(OH)-CH2-CH(C2H5)-OH

fgiga [73]

The IUPAC name for the organic compound CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol

This compound above belong to class of organic compound called the alkanols

<h3>What are organic compounds?</h3>

Organic compounds are compounds which contains carbon and hydrogen only.

Some few classes of organic compounds includes the following:

Alkanes
Alkenes
Alkynes
Alkanols
Alkanals
Alkanones
Esters
Ketones and other

So therefore, the IUPAC name for the organic compound is CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol

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