Ask question

Ask question

All categories

3 years ago

13

A scuba diver knows that she needs 50.0mol of air for an upcoming dive. What size (volume) tank will she need to fill for this d

ive?*

1 answer:

erica [24]3 years ago

4 0

Answer:

1120 L.

Explanation:

Hello!

In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.

In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:

$V_2=\frac{V_1*n_2}{n_1}$

Thereby, we plug in to obtain:

$V_2=\frac{22.4L*50.0mol}{1mol}\\\\V_2=1120 L$

Best regards!

You might be interested in

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

4 0

2 years ago

Which statement describes the use of a flowchart?

castortr0y [4]

Answer: I believe it is B

5 0

3 years ago

Read 2 more answers

Draw the structure of the compound identified by the following simulated 1H and 13C NMR spectra. The molecular formula of the co

Luba_88 [7]

Answer: 4-allylanisole

Explanation: The doublets behind the 7 ppm belongs to the

para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group

A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.

Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.

The Predicted Number is attached from a chemical database along with their peaks information

6 0

3 years ago

Why was Rutherford's Gold Foil Experiment so important?

Oksanka [162]

Answer:

Answer is 'B' I think

Explanation:

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.

4 0

3 years ago

Marking anybody who got it the brainliest

bija089 [108]

I can't open pdf.......

7 0

2 years ago