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2 years ago

What is the approximate partial pressure of oxygen at 1500 m?

Chemistry

1 answer:

elena55 [62]2 years ago

3 0

The approximate partial pressure of oxygen at 1500 m is 0.18 atm.

<h3>Pressure of the air</h3>

From barometric pressure table;

1500 m = 85 kPa

1 kPa = 0.0099 atm

85 kPa = ?

= 0.84 atm

Oxygen percentage in air = 21%

<h3>Partial pressure of oxygen in air</h3>

P = 0.21 x P(total)

P = 0.21 x 0.84 atm

P = 0.18 atm

Thus, the approximate partial pressure of oxygen at 1500 m is 0.18 atm.

Learn more about partial pressure here: brainly.com/question/19813237

#SPJ11

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HELP ME PLEASE I BEG YOU

babunello [35]

True I think is the answer

3 0

3 years ago

A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of

inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 $\frac{g}{c m^{3}}$

To find:

Mass of the copper cylinder

<u>Step by Step by explanation:</u>

Solution:

According to the formula, Mass can be calculated as

$\rho=\frac{m}{v}$ and from this

$m=\rho \times v$

Where, m=mass of the cylinder

$\rho$ =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

$v=\pi r^{2} h$

Where $\pi$ =3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, $v=3.14 \times 12^{2} \times 44$

v=3.14 \times 144 \times 44

$v=19895.04 \mathrm{cm}^{3}$

And we know that, $m=\rho \times v$

Substitute the known values in the above equation we get

$m=8.90 \times 19895.04$

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

4 0

3 years ago

What are the rules for naming hydrocarbons?

GrogVix [38]

Answer:

IUPAC Rules for Alkane Nomenclature

Find and name the longest continuous carbon chain.

Identify and name groups attached to this chain.

Number the chain consecutively, starting at the end nearest a substituent group.

Designate the location of each substituent group by an appropriate number and name.

Explanation:

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2 years ago

When performing physical exercise, the heart rate, by what percentage can it be increased?

Dafna11 [192]

Answer:

Explanation:

I did that already. You got beo

4 0

3 years ago

Read 2 more answers

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

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3 years ago