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3 years ago

K₂O +2Li=>Li₂O +2k ???

Chemistry

2 answers:

Elena L [17]3 years ago

6 0

Potassium + Lithium Oxide = Lithium + Potassium Oxide
Element 19 Kalium Potassium K-38 Potassium K-43 Potassium K-40 Potassium K-42
Molar mass K

Molar Mass of Li2O Oxidation State of Li2O

Element 3 Lithium-6 Atom (6)Li (6)3Li Lithium-6

Potash Potassium Oxidopotassium Dipotassium Oxide Dipotassium Monoxide

Alborosie3 years ago

5 0

Answer:

uhh what even is this. 0_0

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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

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A sample of methane gas has a volume of 10 l at a pressure of 1.5 atm and a temperature of 20 °c. what volume does the gas occup

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We can use combined gas laws to solve for the volume of the gas
$\frac{PV}{T}$
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