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2 years ago

Of the following metals, which is the LEAST reactive? A. K B. Ca C. Na D. Mg

Chemistry

2 answers:

emmainna [20.7K]2 years ago

6 0

The Answer Should Be D.Mg

sladkih [1.3K]2 years ago

3 0

Answer: D. Mg

Explanation:

Chemical reactivity is defined as the tendency of an element to loose of gain electrons.

Metals are the elements which lose electrons and hence, their chemical reactivity will be the tendency to lose electrons.

Their chemical reactivity increases as we move top to bottom in a group because the valence shell come gets far away from the nucleus. Thus, the loss of electron from the valence shell becomes easier due to lesser attraction between nucleus and valence electron.

Their chemical reactivity decreases as we move from left to right in a period .As, the electrons get added up in the same shell, the electron in the outermost orbital gets near to the nucleus. And hence, the electron will be difficult to lose.Thus, the chemical reactivity of metals decreases.

For the given options:

Potassium(K) belongs to Group 1 and period 4 of the periodic table.

Calcium (Ca) belongs to Group 2 and period 4 of the periodic table.

Sodium (Na)belongs to Group 1 and Period 3 of the periodic table.

Magnesium (Mg) belongs to Group 2 and Period 3 of the periodic table.

Order of reactivity of metals follow:

$K>Ca>Na>Mg$

Thus the LEAST reactive metal is Magnesium.

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Convert 6.0cc to cm cubed

ELEN [110]

Cc stands for cm cubed (cubic centimetre).
So
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2 years ago

A standard solution is prepared for the analysis of fluoxymesterone (c20h29fo3; 336 g/mol), an anabolic steroid. a stock solutio

-Dominant- [34]

Answer:

The concentration of new solution will be 1×10∧-7 M.

Solution:

Data Given

given mass of fluoxymesterone =16.8mg = 0.0168 g

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vol. of fluoxymesterone = 500.0 ml = 0.500 L

Stock Molarity of fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M

So applying dilution formula

Stock Solution : New Solution

M1.V1 = M2.V2

( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L

[( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2

1 × 10∧-7 = M2

Result:

The concentration of new solution M2 will be 1 × 10∧-7

3 0

3 years ago

Locations Cost of land Cost of equipment Cost of mining and reclamation Time taken to mine the area 1 $100,000 $10,000 $5,000 pe

mario62 [17]

Answer:

Location 2 will cost the least to the company to mine.

Explanation:

a) Data and Calculations:

Locations Cost of land Cost of Cost of mining Time taken to

equipment and reclamation mine the area

1 $100,000 $10,000 $5,000 per day 30 days

2 $35,000 $6,000 $4,500 per day 45 days

3 $30,000 $7,500 $3,500 per day 120 days

4 $40,500 $8,000 $7,000 per day 65 days

Locations Cost of land Cost of Cost of mining Total

equipment and reclamation Costs

1 $100,000 $10,000 $150,000 ($5,000 * 30) $260,000

2 $35,000 $6,000 $202,500 ($4,500 * 45) $243,500

3 $30,000 $7,500 $420,000 ($3,500 * 120) $457,500

4 $40,500 $8,000 $445,000 ($7,000 * 65) $503,500

7 0

2 years ago

How many grams of P4O10 would be produced when 0.700 mole of phosphorus is burned?

AysviL [449]

We can solve the equation and show the solution below:

Oxygen atomic number is 16.
Phosphorus atomic number is 32.

We have the molecular weight:
Molecular weight = (31*4) + (16*10)
Molecular weight = 284 grams/mol

Solving for the grams:

0.4 mole (for P4) * (1 mol P4O10/1 mol P4) * (284 grams P4O10/1 mole P4O10)
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The answer is 113.6 grams.

8 0

3 years ago

* Need ASAP * Explain the relationship between percent composition, Empirical formula, and molecular formula.

sweet-ann [11.9K]

By dividing the percentage composition with the molar mass of that element we will get the empirical formula. Then using that empirical formula and formula mass we can find the molecular formula.

Explanation:

The chemical properties of any substance are defined obviously by the different types and relative amounts of atoms constituting its primary entities (in case of covalent compounds the primary entities are molecules and ions in the event of ionic compounds).

A percent composition of any compound gives the mass percent of each element present in the compound; in addition to that frequently it is determined experimentally and utilized to derive an empirical formula of any compound. An empirical formula mass of any covalent compound could be comparable with the molar or molecular mass of a compound to acquire a molecular formula.

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3 years ago