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3 years ago

12. Calculate the hydroxide ion concentration of a solution with pH = 10.25. Show all calculations leading to an answer.

Chemistry

2 answers:

Alex787 [66]3 years ago

6 0

PH= log[H3O+]
10.25=log [H3O+]
[H3O+] = 10^10.25
[H3O+]= 1.778 ×10^10

kkurt [141]3 years ago

6 0

<span>pH + pOH = 14 10.25 + pOH = 14 pOH = 14 - 10.25 = 3.75 Now, knowing the pOH we can find the OH^- since pOH is -log [OH-] [OH-] = 10^-3.75 = 1.78x10^-4 M</span>

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Solnce55 [7]

Delta E = Ef - Ei
E = energy , h = plank constant , v = frequency
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E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 ) = 4.90 * 10 ^ -20 J
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3 years ago

How do you find the pressure in a container if you don't have the temperature?

BARSIC [14]

It depends on the process.
Like for example if the process is isothermal(temperature is constant), you can use,
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For adiabatic process,

PV^gamma = constant or P1V1 ^gamma = P2V2 ^gamma.

where gamma = Cp
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Cp = specific heat at constant pressure and Cv = specific at constant volume.

Value of Gamma will be given in question.
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3 years ago

PLEASE HELP NEED THIS TO PASS!!!

nadya68 [22]

(1)

At phase A, the water is fully solid, At phase C the water is fully liquid and at phase E the water is fully gas. At this ponts all the substance is in one state other than the fact that its molecules are getting more excited as they gain more energy. This is why there is an increase in temperatures. At the transition points water is changing phase and therefore at any point in between the transition not all water will be in the same phase.

(2)

Point F

This is because the solid water stops rising in temperatures and begins changing phase into liquid water. This is the highest temperatures that the solid water can attain without turning into liquid. You can see from the graph that the temperatures rise at this point levels off between point F and G

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Point H

This is because the liquid water stops rising in temperatures and begins changing phase into gaseous form. This is the highest temperatures that the liquid water can attain without turning into gas/water vapor. You can see from the graph that the temperatures rise at this point levels off between point H and I.

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The energy is increasing

You can see from the graph that energy is being absorbed by the water (due to increases in energy on the x-axis on the graph) but there is no corresponding rise in temperatures in y-axis. This means the energy begin absorbed is being used to change the phase of the water.

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The water is changing phases

Because the energy absorbed is being used to change the phase of the liquid. The energy is being used to break the bonds between the molecules so the molecules become farther apart and causing a change in phase on the water. Therefore the vibrational moments of the molecules (responsible for rising in temperatures) remain the same over this latent phase. This energy absorbed to change phase is called latent energy/heat.

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3 years ago

8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8

garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7) kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

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3 years ago

If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? ball A,

rewona [7]

Density (D) is defined as the amount or mass (m) of a substance present in a unit volume(V). It can be expressed mathematically as:

Density = Mass/Volume

i.e. D = m/V -------(1)

Units: g/cm3

If m1, V1 and D1 = mass, volume and density respectively of ball A

m2, V2 and D2 = mass, volume and density respectively of ball B

It is given that: V1 = V2; but m1 = 2m2

Based on equation (1) we have:

D1/D2 = (m1/V1)* (V2/m2) = (2m2/V2)*(V2/m2) = 2

Thus, density of ball A is twice that of B.

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If m1, V1 and D1 = mass, volume and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

It is given that:

V1 = 3V2 and m2 = 1/3(m1) i.e m1 = 3m2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (3m2/3V2)*(V2/m2) = 1

Thus, D1 = D2

Ans: Their densities are equal

If m1, V1 and D1 = mass, volume and density respectively of ball P

m2, V2 and D2 = mass, volume and density respectively of ball Q

It is given that:

m1 = m2 but V1 = 2V2

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = (m2/2V2)*(V2/m2) = 1/2

Thus, D1 = 1/2(D2)

Ans: Ball Q will have a greater density.

If m1, V1 and D1 = mass, volume and density respectively of ball X

m2, V2 and D2 = mass, volume and density respectively of ball Y

It is given that:

V1 = 2V2 and m1 = 1/2(m2)

Therefore,

D1/D2 = (m1/V1)* (V2/m2) = ((1/2(m2)/2V2)*(V2/m2) = 1/4

Thus, D1 = 1/4(D2)

Ans: Ball Y will have a greater density.

5 0

3 years ago

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