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horrorfan [7]
3 years ago
12

What is the freezing point (in °C) of a 0.743 m

Chemistry
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

man i sure wish u read the explanation and all of it

Explanation:

We're no strangers to love

You know the rules and so do I

A full commitment's what I'm thinking of

You wouldn't get this from any other guy

I just wanna tell you how I'm feeling

Gotta make you understand

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

We've known each other for so long

Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

And if you ask me how I'm feeling

Don't tell me you're too blind to see

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

(Give you up)

(Ooh) Never gonna give, never gonna give

(Give you up)

We've known each other for so long

Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

I just wanna tell you how I'm feeling

Gotta make you understand

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

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A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
The mass of an erlenmeyer flask is 85.135 g. after 10.00 ml of water is added to the flask, the mass of the flask and the water
BARSIC [14]
Subtracting the mass of (flask+water) from the empty flask gives:
95.023 g - 85.135 g = 9.888 grams of water
Dividing this by the given volume of 10.00 mL water gives:
9.888 grams of water / 10.00 mL of water = 0.9888 g/mL of water
Therefore, based on this sample, the density of water is 0.9888 g/mL, which is close to the usually accepted approximation of 1 g/mL.
4 0
3 years ago
What property do atoms of these elements have that helps make the molecules polar
DENIUS [597]
You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
even distribution -------> non-polar
4 0
3 years ago
How many grams of K2CO3 are needed to prepare a 400.0 mL of a 4.25M solution?
Basile [38]

Answer:

235 g

Explanation:

From the question;

  • Volume is 400.0 mL
  • Molarity of a solution is 4.25 M

We need to determine the mass of the solute K₂CO₃,

we know that;

Molarity = Number of moles ÷ Volume

Therefore;

First we determine the number of moles of the solute;

Moles = Molarity × volume

Moles of  K₂CO₃ = 4.25 M × 0.4 L

                           = 1.7 moles

Secondly, we determine the mass of  K₂CO₃,

We know that;

Mass = Moles × Molar mass

Molar mass of  K₂CO₃, is 138.205 g/mol

Therefore;

Mass = 1.7 moles × 138.205 g/mol

         = 234.9485 g  

         = 235 g

Thus, the mass of  K₂CO₃ needed is 235 g

7 0
3 years ago
HELPPP which one has four energy levels Si, Ge, or Sn???
vaieri [72.5K]

Answer:

Ge

Explanation:

pretty sure im on high school 12th

8 0
3 years ago
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