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vlabodo [156]
3 years ago
8

What is common about the elements of a period? Select all that apply. They have the same number of electron energy levels. They

have similar chemical characteristics. They have similar physical characteristics. They transition from a metal to noble gas.
Chemistry
1 answer:
Ann [662]3 years ago
4 0

Answer:

They have the same number of electron energy levels.

They transition from a metal to noble gas.

Explanation:

Periods in the periodic table of elements refer to elements in the same row. All the elements in a certain row of the periodic table;

have the same number of electron energy levels.

transition from a metal to noble gas.

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Which process is used to produce gases from solutions of salts dissolved in water or another liquid?
NemiM [27]

Answer:

Explanation:

Option D electrolysis is the correct answer

3 0
3 years ago
Read 2 more answers
Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction
Luden [163]

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

<u />

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

<em>The chemical equation for this reaction is:</em>

HA  +  B  ⇌  A⁻  +  HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Reason: In this reaction, there is no exchange of proton between the acid and the base.

4 0
2 years ago
A mass of 100.0 g of NaCl is added to 100.0 mL of water, but not all of it dissolves. A mass of 59.5 grams of NaCl solid remains
Taya2010 [7]

Answer:

The molarity of the dissolved NaCl is 6.93 M

Explanation:

Step 1: Data given

Mass of NaCl = 100.0 grams

Volume of water = 100.0 mL = 0.1 L

Remaining mass NaCl = 59.5 grams

Molar mass NaCl= 58.44 g/mol

Step 2: Calculate the dissolved mass of NaCl

100 - 59. 5 = 40.5 grams

Step 3: Calculate moles

Moles NaCl = 40.5 grams / 58.44 g/mol

Moles NaCl = 0.693 moles

Step 4: Calculate molarity

Molarity = moles / volume

Molarity dissolved NaCl = 0.693 moles / 0.1 L

Molarity dissolved NaCl = 6.93 M

The molarity of the dissolved NaCl is 6.93 M

8 0
2 years ago
Read 2 more answers
Why should reactions not be performed in graduated cylinder?​
Stells [14]

Answer:

A Graduated Cylinder has a limited round and hollow shape with each stamped line demonstrating the volume of fluid being estimated. While they are commonly more exact and exact than lab flagons and recepticles, they ought not be utilized to perform volumetric examination.

3 0
3 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
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