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2 years ago

What electives should i take in highschool to prepare to take criminology?

Chemistry

1 answer:

dybincka [34]2 years ago

4 0

Answer:

depending on what school you go to and what classes you are allowed to take in which grades, you should take forensics.

You might be interested in

In reversible reaction when do forward reactions take place

kari74 [83]

Answer:

that is true. because reversible react is take place in forward and backward reaction

7 0

3 years ago

A travel mug of 91∘C coffee is left on the roof of a parked car on a cold winter day. The temperature of the coffee after t minu

GREYUIT [131]

Answer: A

Explanation:

3 0

3 years ago

Alkaline earth metals are very reactive because of how many valence electrons ?

Alborosie

Their is 2 valence electrons

8 0

3 years ago

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete

Katarina [22]

We will use boiling point formula:

ΔT = i Kb m

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass Kg

0.27 = moles / 0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol

5 0

3 years ago

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

<u>Answer:</u> The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

3 0

3 years ago