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NemiM [27]
3 years ago
9

A typical air sample in the lungs contains oxygen at 100 mmhg, nitrogen at 573 mmhg, carbon dioxide at 40 mmhg, and water vapor

at 47 mmhg. part a why are these pressures called partial pressures
Chemistry
1 answer:
nikdorinn [45]3 years ago
4 0
  it  is  called the  partial pressure   because    it  is    pressure  that   is  exerted by  each gas in  a mixture of  gases if it occupy  the same   volume  of  it  own.  the  the partial pressure  of oxygen  in the air sample above is  100 mm hg, that of  carbon dioxide is 40 mmhg, for   water vapor  47 mmhg. and that of  nitrogen  is 573 mmhg
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Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of
jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = 2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x

2330.35x=1048815\\x=450.068g

Hence, there was 450.068g of water in the pot.

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How do you represent a shared pair of electrons when drawing the Lewis structure of a covalent bond
nataly862011 [7]

Answer: Each pair of shared electrons is a covalent bond which can be represented by a dash.

Explanation:

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You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

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v_2 = actual volume of water = 100-mL

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∴

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\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

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