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3 years ago

Determine the pH of a 2.8 ×10−4 M solution of Ca(OH)2.

Chemistry

1 answer:

shepuryov [24]3 years ago

3 0

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

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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.

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4 years ago

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3 years ago

What is the mass of a sample of alcohol (specific heat = 2.4 J/gC), if it requires 4780 J of heat to raise the temperature by 5.

insens350 [35]

The mass of a sample of alcohol is found to be = m = 367 g

Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.

Explanation:

The Energy of the sample given is q = 4780

We are required to find the mass of alcohol m = ?

Given that,

The specific heat given is represented by = c = 2.4 J/gC

The temperature given is ΔT = 5.43° C

The mass of sample of alcohol can be found as follows,

The formula is c = $\frac{q}{mt}$

We can drive value of m bu shifting m on the left hand side,

m = $\frac{q}{ct}$

mass of alcohol (m) = $\frac{4780}{(2.4)( 5.43)}$

m = 367 g

Therefore, The mass of the given sample of alcohol is

m = 367g

It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.

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4 years ago

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of $4.71\times 10^{-15}J$ . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

= De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of beta particle = $9.1094\times 10^{-31} kg$

$E_k$ = kinetic energy of the particle = $4.71\times 10^{-15}J$

Putting values in above equation, we get:

$\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}$

$\lambda = 6.762\times 10^{-12} m$

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

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3 years ago

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kogti [31]

Answer: The steepness of a ramp affects it by making it easier or harder.

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3 years ago