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Alecsey [184]
2 years ago
15

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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In a certain time, light travels 4.96 km in a vacuum. during the same time, light travels only 3.36 km in a liquid. what is the
Natali5045456 [20]

the refractive index of the liquid is 1.476

The refractive index, which has no dimensions, measures how quickly light passes through a substance.

It can also be described as the difference between the speed of light in a vacuum and a medium.

Refractive index is equal to the product of the light's liquid and vacuum speeds.

Therefore.

speed of light in vacuum = 4.96 km/t

speed of light in liquid = 3.36 km/t

Refractive index = 4.96/3.36

Refractive index =1.476

Therefore, the refractive index of the liquid is 1.476

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brainly.com/question/23750645

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6 0
1 year ago
How many particles are in 5.00 grams of nitric acid?
Leto [7]

Answer:

0.0500

Explanation:

7 0
2 years ago
An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
RSB [31]

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

Best regards!

7 0
3 years ago
What is the osmotic pressure, in torr, of a 3.00% solution of NaCl in water when the temperature of the solution is 45 ºC? Enter
Anit [1.1K]

213034 torr is the osmotic pressure.

Explanation:

osmotic pressure is calculated by the formula:

osmotic pressure= iCrT

where i= no. of solute

c= concentration in mol/litre

R= Universal Gas constant

T = temp

It is given that solution is 3% which is 3gms in 100 ml.

let us calculate the concentration in moles/litre

3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl

= 5.372 gm/litre

Putting the values in the formula,                   Temp in Kelvin 318.5K

osmotic pressure= 2*5.372*0.083 * 318.5                 Gas constant  0.083

                              = 284.023 bar or 213018 torr.                 c=  5.372 moles/L                                                                              

                                                                                    i=2 for NaCl

4 0
3 years ago
Which scientific law states the relationship between an object's mass, acceleration, and amount of force acting on it? *
jenyasd209 [6]
Newton's second law of motion
5 0
3 years ago
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