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vredina [299]
3 years ago
13

Please fast anwser ill give you points!!! TY

Chemistry
1 answer:
Misha Larkins [42]3 years ago
8 0
 I think it's D. helps to analyze the experiment because the purpose is to show how you got the conclusion that you did and how to keep track of any mistakes you might have made along the way.
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If the accepted value for the mass of an object is 20.0g and the student found that the mass of the object was 20.5g what is the
NikAS [45]

Answer:

Results

The percent error between 20 and 20.5 is 2.5%

Explanation:

Percent Error = | (20.5 − 20) / 20 | × 100 = | (0.5) / 20 | × 100 = | 0.025 | × 100 = 2.5% (three decimal places)Percent Error = 2.5%

3 0
3 years ago
Which electron configuration is correct for a sodium ion?
jeka94

The electron configuration for sodium ion Na⁺ : <u>(2) 2-8</u>

<h3>Further explanation </h3>

In an atom there are levels of energy in the shell and sub shell

This energy level is expressed in the form of electron configurations.

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d and f. The maximum number of electrons for each sub shell is

  • s: 2 electrons
  • p: 6 electrons
  • d: 10 electrons and
  • f: 14 electrons

Charging electrons in the sub shell uses the following sequence:

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc. </em>

Determination of electron configurations based on principles:

• 1. Aufbau: Electrons occupy orbitals of the lowest energy level

• 2 Hund: electrons fill orbitals with the same energy level

• 3. Pauli: no electrons have the same 4 quantum numbers

The alkali metal Na will release electrons to form Na + so that the electron configuration is stable as the noble gas element Ne

electron configuration Ne: [He] 2s² 2p⁶

Na electron configuration: [Ne] 3s¹

electron configuration Na + = [Ne] = [He] 2s² 2p⁶

The maximum number of electrons in the shell K, L, M, N

According to Bohr, the maximum number of electrons that can occupy each atomic shell can be calculated by the formula 2n²

  • K shell (n = 1): 2.1² = 2 electrons
  • L shell (n = 2): 2. 2² = 8 electrons
  • M shell (n = 3): 2. 3² = 18 electrons
  • N shell (n = 4): 2. 4² = 32 electrons

If we look at the configuration of the Na⁺ ion:

1s² 2s² 2p⁶ then

on the shell n=1 (1s) there are 2 electrons

on the shells n=2 (2s and 2p) there are 8 electrons

So that the configuration

Na⁺ = 2-8

<h3>Learn more </h3>

element X

brainly.com/question/2572495

electrons and atomic orbitals

brainly.com/question/1832385

Identify the group number in the periodic table

brainly.com/question/2014634

5 0
3 years ago
Read 2 more answers
A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac
UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

pH = pKa + log(\frac{[KF]}{[HF]})

pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol

\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles

And the concentration of HF after the reaction of KF with HCl is is:

C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L

Now, from the equilibrium of equation (1) we have:

Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}

Ka = \frac{x^{2}}{0.531 - x}  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

8 0
4 years ago
An element has this information:
SVETLANKA909090 [29]

Answer:

hydrogen

Explanation:

5 0
3 years ago
1. How many GRAMS of potassium nitrate are present in 4.65 moles of this compound?
strojnjashka [21]
Basically for the first question, it’s 470.58 grams of potassium nitrate. Then for the second, it’s 0.34 moles of potassium nitrate. I hope the work makes sense!

3 0
3 years ago
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