Consider the reaction.

WILL MARK YOU THE BRAINLIEST IF YOU ANSWER THESE TWO QUESTION CORRECTLY :)

lara [203]

Answer: 1. HYDROCARBONS? 2. ALKANES?

i'm not exactly AMAZING at this but i did some research and this is what i think it is i'm also not in this grade but i tried.

7 0

3 years ago

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

0.055 grams of PbSO4 is dissolved in 200.00 grams of H2O.

Greeley [361]

Answer:

Solution's mass = 200.055 g

[PbSO₄] = 275 ppm

Explanation:

Solute mass = 0.055 g of lead(II) sulfate

Solvent mass = 200 g of water

Solution mass = Solvent mass + Solution mass

0.055 g + 200 g = 200.055 g

ppm = μg of solute / g of solution

We convert the mass of solute from g to μg

0.055 g . 1×10⁶ μg/ 1g = 5.5×10⁴μg

5.5×10⁴μg / 200.055 g = 275 ppm

ppm can also be determined as mg of solute / kg of solution

It is important that the relation is 1×10⁻⁶

Let's verify: 0.055 g = 55 mg

200.055 g = 0.200055 kg

55 mg / 0.200055 kg = 275 ppm

6 0

3 years ago

Which is most often used to destroy bacteria?

vovangra [49]

Iodine- in most cleaning solutions

7 0

3 years ago

Write the net ionic equation for the reaction between hydrochloric acid and potassium cyanide. Do not include states such as (aq

zhannawk [14.2K]

Answer: $H^++CN^-\rightarrow HCN$ , product favoured

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we do not include the spectator ions in the equations.

When hydrochloric acid react with potassium cyanide, then it gives potassium chloride and hydrocyanic acid as products.

The complete ionic equation will be:

$H^++Cl^-+K^++CN^-\rightarrow K^++Cl^-+HCN$

The net ionic equation will not contain spectator ions which are $K^+$ and $Cl^-$ :

$H^++CN^-\rightarrow HCN$

The reaction is product favoured.

5 0

3 years ago