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2 years ago

DDT is currently banned worldwide. a. True b. False

1 answer:

5 0

B. False. It is banned worldwide for agricultural uses. The use of DDT is still authorized in small proportions in countries that need it, with support deployed for the alteration to safer and more effective alternatives.

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What is exchanged between the system and the surroundings

BartSMP [9]

Answer:

see note under explanation

Explanation:

When describing system and surroundings the system is typically defined as the 'object of interest' being studied and surroundings 'everything else'. In thermodynamics heat flow is typically defined as endothermic or exothermic. However, one should realize that the terms endothermic and exothermic are in reference to the 'system' or object of interest being studied. For example if heat is transferred from a warm object to a cooler object it is imperative that the system be defined 1st. So, with that, assume the system is a warm metal cylinder being added into cooler water. When describing heat flow then the process is exothermic with respect to the metal cylinder (the system) but endothermic to the water and surroundings (everything else).

8 0

3 years ago

Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)

miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

Adding solute to water elevates the boiling point.

The elevation in boiling point (ΔTb) can be calculated using the relation:

ΔTb = i.Kb.m,

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

(1) 1.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

(2) 2.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

(3) 1.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

(4) 2.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).

6 0

3 years ago

The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is __________

PIT_PIT [208]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl

Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality = 0.7543 mol / kg

The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is 0.7453 molal.

6 0

2 years ago

A 143.1 g sample of a compound contains 53.4 g of carbon, 16.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. what

nika2105 [10]

I'm fairly sure this is it 20.82%

6 0

3 years ago

Read 2 more answers

At what temperature (in C) will a sample of gas occupy 91.3 L if it occupies 45.0 L at 70.0°C? Assume constant pressure.)

sdas [7]

Solution is here,

for initial case,

temperature(T1)=70°C=70+ 273=343K

vloume( V1) =45 L

for final case,

temperature( T2)=?

volume(V2)= 91.3 L

at constant pressure,

V1/V2 = T1/T2

or, 45/91.3 = 343/ T2

or, T2= (343×91.3)/45

or, T2=695.9 K = (695.9-273)°C=422.9°C

5 0

2 years ago