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3 years ago

Describe two shortcomings of the pennium model for isotopes

1 answer:

3 0

let me know when u find out plz because i would like to know as well its one of my chemistry qustions in an assiment. :)

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Imagine you are given a mystery element. It is, however, a discovered and known element. You may perform a maximum of two observ

miv72 [106K]

The classification of it being a metal, nonmetal, or metalliod will be useful in the process of elimination to determine what it is. Then for the second test, meauring the atomin radius will narrow it down quicker to the mystery elemet's name.

Since you determined what part of the periodic table it's on, then when measuring the atomic radius, you should be able to pinpoint what the element is more surely.

6 0

2 years ago

Read 2 more answers

Name the functional group in the

8_murik_8 [283]

It’s called Diethyl ether so I think the answer is D

8 0

2 years ago

Explain the digestive process in an essay.

fiasKO [112]

Refer to attachment for your answer

8 0

1 year ago

How is the concept of heat related to the kinetic molecular theory and the law of conservation of energy?

Slav-nsk [51]

The first law states that the total increase in the energy of a system is equal to the increase in thermal energy, meaning that heat is a form of energy and is therefore subject to the the principle of conservation

7 0

2 years ago

An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that $pH = -log[H^{+}]$ , and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

$pH = pKa - log \frac{[HA]}{[A^{-}]}$

Where [HA] is the concentration of the acid, and $[A^{-}]$ is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of $OH^-$ will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of $OH^-$ reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of $OH^-$ , which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

$6.72 = pKa - log \frac{0.01}{0.0025}$

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0

3 years ago