What is the density of an apple

Which of the following is not a part of kinetic molecular theory?

Gala2k [10]

C. When temperature increase, particles’ speed is suppose to increase. It comes from the ideal gas law.

4 0

3 years ago

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For the function F(x) = 1, which of these could be a value of F(x) when x is<br> close to zero?

Sonbull [250]

Answer:

A. -10,000

Explanation:

When x approaches zero, the graph tells us that F(x) either becomes extremely large or extremely small. The best answer is therefore A: -10,000. Best, because we aren't given the exact value of x, but clearly the other results require that x be greater than 1 or less than -1, which is far from zero.

7 0

2 years ago

Phosphoric acid (H3PO4) is a triprotic acid because each of its three hydrogen atoms may dissociate in water. What mathematical

Juli2301 [7.4K]

The formula that connects Ka1, Ka2, and Ka3 is the relation;

Ka = Ka1 * Ka2 * Ka3

The Ka is defined as the acid dissociation constant of a solution. Now we have the triprotic acid called phosphoric acid. this implies that we have Ka1, Ka2, and Ka3.

Thus the formula that connects Ka1, Ka2, and Ka3 is the relation;

Ka = Ka1 * Ka2 * Ka3

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6 0

2 years ago

What will be the effect on the solubility of BaF2 if the following changes are made: increasing temperature; adding NaF; adding

borishaifa [10]

In BaF₂ the solubility will decreases on adding NaF

the solubility will increases on adding HCl

<h3>SOLUBILITY OF BARIUM FLUORIDE</h3>

Increase in temperature will increase the solubility of the solid.
By adding NaF - decrease the solubility of BaF₂
By adding HCl - increase the solubility of BaF₂

<h3>BARIUM FLUORIDE</h3>

It is colourless solid that occur as rare mineral
It is corroded by moisture
It is used in window of IR spectroscopy

Hence the barium fluoride the solubility decreases on adding NaF and increases on adding HCl.

Learn more about the solubility on

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3 0

2 years ago

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

Temka [501]

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(c) n = 2 to n = 5

(d) n = 4 to n = 3

Answer:

Hence the order of the transition will be : d < a < c < b

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = 2 in an hydrogen atom:

$E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV$

Energy of n = 3 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy of n = 4 in an hydrogen atom:

$E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV$

Energy of n = 5 in an hydrogen atom:

$E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV$

a) n = 2 to n = 4 (absorption)

$\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV$

b) n = 2 to n = 1 (emission)

$\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV$

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

$\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV$

d) n = 4 to n = 3 (emission)

$\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV$

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

$E=h\nu$

h = Planck's constant

$\nu$ frequency of the wave

So, the increasing order of magnitude of the energy difference :

$E_4$

And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :

: d < a < c < b

7 0

4 years ago