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wariber [46]
3 years ago
9

Sodium is going in group 1, period 3. It contains A. One valence electron in its 3rd energy shell B three valence electron in it

s 1st energy shell C. 11 protons, 11electrons , 11 neutrons D. 11 protons, 11 electrons and 23 neutrons
Chemistry
1 answer:
topjm [15]3 years ago
5 0

Answer:

The correct answer is option A, that is, one valence electron in its third energy shell and option C, that is, 11 electrons and 11 protons.

Explanation:

The outermost electrons and the ones that take part in the process of bonding are termed as valence electrons. The atomic number of sodium is 11, thus, it possesses 11 protons and the atoms are neutral so it suggests that sodium has 11 electrons. By electronic configuration, it can be seen that in sodium, two electrons are present in the first shell, 8 in the second, and only one electron in the third shell, that is, 2.8.1. The electron present in the third shell is the valence electron.

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What is the ph of a saturated solution of aluminum hydroxide?
sasho [114]
Answer is attached  below ( sorry for mistake )

6 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
ololo11 [35]
Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995

So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

Have a nice day!
3 0
3 years ago
Joseph Proust showed that when elements combine to form new substances, they do so in specific mass ratios.a) trueb) false
olga55 [171]

Answer: option A) True

Explanation:

Joseph Proust is the founder of the Law of Definite proportion. This law states that two elements will always combine together to form a chemical compound and this will be in the same proportion by mass.

For Example: 2 moles of Hydrogen (H2) ALWAYS combine with 1 atom of Oxygen (O) to yield water (H2O)

So, the answer is True

3 0
3 years ago
If 50.0 g of formic acid (hcho2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of soluti
gizmo_the_mogwai [7]

If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

<h3>To calculate concentration of sodium formate and formic acid</h3>

The ratio of number of moles and the volume of solution is Molar concentration of substance.

Concentration of sodium formate

Cb = 30/(68×500)

= 0.00088m

Concentration of formic acid

Ca = 50/(46×500)

= 0.00217m

Now,

by using Henderson hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

= 3.75

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75 - 0.392

pH = 3.35

Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.

learn more about pH:

brainly.com/question/13423434

#SPJ4

3 0
2 years ago
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