How does the electric force between two charged particles change if the

Name an element or elements in the periodic table that you would expect to be chemically similar to potassium. check all that ap

Inga [223]

It would have to be all elements in the same family, so since potassium is an Alkali Metal the similar chemicals would be...Lithium, Sodium, Rubidium, Cesium, Francium.

7 0

3 years ago

By what factor does [h+ ] change for each ph change? (a) 3.20 units

konstantin123 [22]

When a change in PH = 10^-ΔPH
so the change = 10^-3.2
change depends on two factor 0.00063 (10^-3.2)and factor 1585 (10^3.2)depending on the way which the change goes.if PH change from PH=0 to PH= 3.2 so the change is decreasing from concentration from 1 mol to 0.00063 and if PH change from PH = 3.2 to PH=0 so the change is increasing by a factor of 1585.

3 0

3 years ago

Please help me for question 1 and 2

likoan [24]

Answer:-

1) 6 mol

2) Mo

Explanation: -

Mass of Ozone = 48 g

Chemical formula of ozone = O3

Molar mass of Ozone O 3 = 16 x 3 = 48 g mol-1

Number of moles of ozone = Mass / molar mass

= 48 g / 48 g mol-1

= 1 mol

According to Avogadro’s law, 1 mole of a substance has 6.02 x 10^ 22 molecules.

So 1 mol of O3 has 6.02 x 10^ 22 molecules of ozone.

Now each Ozone molecule has 3 atoms of oxygen.

So, 1 mol of ozone has 3 x 6.02 x 10^22 atoms of oxygen.

Sodium must have 2 x 3 x 6.02 x 10^22 atoms as per the question.

According to Avogadro’s law, 6.02 x 10^ 22 atoms are in 1 mol of sodium

So, for 2 x 3 x 6.02 x 10^22 atoms, there should be (1/ 6.02 x 10^ 22) x 2 x 3 x 6.02 x 10^22

= 6 mol of sodium.

b)

Let the mass of M be m g

Formula of hexafluoride = MF6.

Mass of the hexafluoride = g + 6 x 19

= m + 114

Mass of M=0.250g

Moles of M = 0.250/m

Mass of MF6= 0.547g

Moles of MF6 = 0.547/ (m + 114)

We know 1 mole of M gives 1 mole of MF6.

0.250/m moles of M gives 0.250/m moles of MF6.

But number of moles of MF6 = 0.547/ (m + 114)

Thus

0.250/m = (0.547)/ (m +114))

0.250m + 0.250 x 114 = 0.547m

m = 0.250 x 114 / (0.547 -0.250)

= 96

We see from the given data that Mo is 96.

So M is Mo.

4 0

3 years ago

What is an input force?

nataly862011 [7]

Force is a pull or a push acting on a body at rest or in motion resulting from its interaction with another body. Input force is the force that you put on a machine while Output force is the force the machine exerts on an object. The output distance is when the output force moves the machine a certain distance while the input distance is when the input distance is when the input force moves the machine a certain distance.

6 0

4 years ago

Read 2 more answers

100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.

wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

5 0

3 years ago