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3 years ago

Which of these contains generic material but is not classified as living?

Chemistry

1 answer:

denis-greek [22]3 years ago

4 0

Viruses are not classified as being alive in because of the fact that they can't reproduce.

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Why do farmers spray water over their crops before a frost?

Nezavi [6.7K]

Answer:

When water freezes and turns into ice, it releases latent heat. Then, the ice that builds up on the plant will insulate it from the colder surrounding air temperatures. Because of this, some growers choose to spray their crop with water before the freeze occurs.

Explanation:

5 0

3 years ago

What effect does increasing the concentration of a dissolved solute have on each of the colligative properties?

Igoryamba

Answer:

If we increase the concentration of a dissolved solute, the solution would have a vapor pressure so much low, the boiling temperature for the solution will be so high, freezing point for the solution will be so much low and the osmotic pressure will be higher.

Colligative properties always depends on dissolved particles (solute)

Explanation:

These are the colligative properties

- Vapor pressure lowering

ΔP = P° . Xm

Vapor pressure of pure solvent - Vapor pressure of solution.

If we add more solute, it would raise the Xm, so the solution would have a vapor pressure so much low.

Vapor pressure pure solvent - Vapor pressure solution ↑ = P° . Xm ↑

- Boiling point elevation

ΔT = Kb . m

When we add more solute, we are increasing the molality.

↑T° boiling of solution - T° boiling pure solvent = Kf . m ↑

Boiling temperature for the solution will be so high.

- Freezing point depression

When we add more solute, we are increasing the molality.

ΔT = Kf . m

T° fussion of pure solvent - ↓T° fussion of solution = Kf . m↑

Freezing point for the solution will be so much low.

- Osmotic pressure

π = M . R . T

When we add solute, molarity is increasing. Therefore the osmotic pressure will be higher.

π↑ = M↑ . R . T

7 0

3 years ago

At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.

Goryan [66]

Answer:

Kc = 2.34 mol*L

Explanation:

The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.

A + B ⇄ C + D

Kc = [C] * [D] / [A] * [B]

According to the reaction

Kc = [SO2]^2 * [O2]^2 / [SO3]^2

Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3

0.450 --> 0 + 0 (Beginning of the reaction)

0.260 --> 0.260 + 0.130 (During the reaction)

0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)

Kc = [0.260]^2 + [0.130]^2 / [0.190]^2

Kc = 2.34 mol*L

3 0

3 years ago

Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo

Luda [366]

<u>Answer:</u> The acceleration of the object is $2m/s^2$ . If net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

<u>Explanation:</u>

Force is defined as the product of object's mass and acceleration.

Mathematically,

$F=ma$ ......(1)

or,

$a=\frac{F}{m}$ .....(2)

where,

F = Force exerted on an object = 60N

m = mass of an object = 30kg

a = acceleration of the object = ?

Putting values in above equation, we get:

$a=\frac{60kg.m/s^2}{30kg}=2m/s^2$

The acceleration of the car is $2m/s^2$ .

From equation 2, it is visible that acceleration is directly proportional to force. This means that \if force increases, acceleration also increases.

And acceleration is inversely proportional to mass of the object. This means that if mass increases, the acceleration decreases.

Hence, if net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

8 0

3 years ago

Help plzzzzzzzzzzzzzzzzzz

alexgriva [62]

Answer:

1.c

2.e

3.a

4.b

5.d

7 0

3 years ago