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3 years ago

What happens on a molecular level when a diatomic molecule is a gas but is then cooled to a solid?

Chemistry

1 answer:

irina1246 [14]3 years ago

4 0

Answer:

As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass.

Explanation:

Hopefully that helps!

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How would using a wider cuvette affect absorbance?

IgorC [24]

The width of the cuvette is important in the spectrophotometric analysis since wider cuvette translates to more absorbing species present in the path where light passes, hence more absorbance is read in the analysis. other factors include the concentration of the sample and the species itself present.

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3 years ago

If the pOH increases is a solution getting more basic or less basic?

sveticcg [70]

Answer:

Increases

Explanation:

Both acids and bases can measured

using the pH or pOH scale. Both

scales provide a measure of either

the H+ concentration or the OHconcentration.

Notice that each scale shows were

acids and bases both are located.

• When acids are measured, the

pH is less than 7, but the pOH

is greater than 7.

• When bases are measured,

the pH is greater than 7, but

the pOH is less than 7.

Both scales are dependent on what

ion you are measuring

6 0

3 years ago

How do you draw the structure of 5-iodo-2-methoxyhexane?

monitta

The structure is in attachment.
Hexane is alkane (acyclic saturated hydrocarbon, carbon-carbon bonds are single) of six carbon atoms. Methoxy group is the functional group consisting of a methyl group bound to oxygen. Iodo is substituent consists of ionide (element in 17 periodic group).

Download docx

8 0

3 years ago

Read 2 more answers

This is the question

iragen [17]

None of these because it doesn’t make since

5 0

2 years ago

Read 2 more answers

16.78 A buffer is prepared by mixing 525 mL of 0.50 M formic acid, HCHO2, and 475 mL of 0.50 M sodium for- mate, NaCHO2. Calcula

dolphi86 [110]

Answer: Before addition of HCl, the pH is 3.70 and after addition of HCl, the pH is 3.64 .

Explanation: The pH of the buffer solution is calculated by using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

Let's calculate the moles of acid and base(salt) present in the original buffer.

mL are converted to L and then multiplied by molarity to get the moles.

$525mL(\frac{1L}{1000mL})(\frac{0.50molHCHO_2}{1L})$

= $0.2625molHCHO_2$

$475mL(\frac{1L}{1000mL})(\frac{0.50molNaCHO_2}{1L})$

= $0.2375molNaCHO_2$

Total volume of the buffer solution = 0.525 L + 0.475 L = 1.00 L

Since, the total volume is 1.00 L, concentration of base will be 0.2375 M and the concentration of acid will be 0.2625 M.

pKa for formic acid is 3.74. Let's plug in the values in the equation and calculate the pH of the original buffer.

$pH=3.74+log(\frac{0.2375}{0.2625})$

pH = 3.74 - 0.04

pH = 3.70

Now, we add 8.6 mL of 0.15 M HCl acid to 85 mL of the buffer. Let's calculate the moles of acid and base in 85 mL of the buffer.

$85mL(\frac{1L}{1000mL})(\frac{0.2625molHCHO_2}{1L})$

= $0.0223molHCHO_2$

$85mL(\frac{1L}{1000mL})(\frac{0.2375molNaCHO_2}{1L})$

= $0.0202molNaCHO_2$

Now, let's calculate the moles of HCl added to the buffer.

$8.6mL(\frac{1L}{1000mL})(\frac{0.15molHCl}{1L})$

= 0.00129 mol HCl

This added HCl reacts with base(sodium formate) and formic acid is produced.

So, 0.00129 moles of HCl will react with 0.00129 moles of sodium formate to produce 0.00129 moles of formic acid. We can write formate ion in place of sodium formate and hydrogen ion in place of HCl. The equation would be:

$H^++CHO_2^-\rightarrow HCHO_2$

moles of base after reaction with HCl = 0.0202 mol - 0.00129 mol = 0.01891 mol

moles of acid after addition of HCl = 0.0223 mol + 0.00129 mol = 0.02359 mol

Let's plug in the values again in the Handerson equation and calculate the pH:

$pH=3.74+log(\frac{0.01891}{0.02359})$

pH = 3.74 - 0.096

pH = 3.64

8 0

4 years ago