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Dvinal [7]
3 years ago
13

Models may change as new discoveries are made. Please select the best answer from the choices provided T F

Chemistry
1 answer:
Serjik [45]3 years ago
8 0
True.
For an example, look at the models of the atom. First, we have the plumb pudding model. Then that was replaced by the planetary model. That model was replaced by the quantum model. 
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MrRissso [65]

Answer:

Any of the answers given will work

Explanation:

I literally just did it.

7 0
3 years ago
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Can someone please help I need help?!?!
anzhelika [568]

Answer:

a: 6 moles

b: 6.75 moles

c: 5 grams

Explanation:

a: mole ratio 2:3

b: mole ratio 2:3

c: mole ratio 2:2

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3 years ago
The mass number of an element is equal to
klasskru [66]

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8 0
3 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
An unknown gas diffuses 1.25 times faster than CO2 gas. What is the molar mass of the unknown gas?​
hodyreva [135]

Answer:

28.16 g/mol

Explanation:

From Graham's law;

Let the rate of diffusion of gas X be 1.25

Let the rate of diffusion of CO2 be 1

Molecular mass of gas X= M

Molecular mass of CO2 = 44g/mol

1.25/1=√44/M

(1.25/1)^2 = 44/M

1.5625 = 44/M

M= 44/1.5625

M= 28.16 g/mol

4 0
3 years ago
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