All categories

3 years ago

The most active metals on the periodic table are found on the: lower left upper right middle upper left lower right

Chemistry

2 answers:

loris [4]3 years ago

7 0

Answer: Option (a) is the correct answer.

Explanation:

Elements which are most electropositive or electronegative are the most reactive in nature. Therefore, they are active.

Elements of group 1 are also known as alkali metals. Elements of group 1 are lithium, sodium, potassium, rubidium, cesium, and francium.

All these are highly electropositive in nature and thus, are active. These are placed at the top left column of periodic table.

Thus, we can conclude that out of the given options, the most active metals on the periodic table are found on the lower left.

andreev551 [17]3 years ago

4 0

Answer: The correct answer is lower left.

Explanation:

Metals are defined as the elements which loose electrons to attain stable electronic configuration.

Reactivity of metals is defined as the tendency of metals to loose electrons. If a metal looses electrons easily, it means that it is more reactive.

Reactivity increases down the group and decreases across a period for metals.

In the periodic table attached,

The elements present in lower left are metals which are most reactive.
The elements present in upper right are non-metals which are most reactive.
The elements present in middle are metals which are less reactive.
The elements present in upper left are metals which are least reactive reactive.
The elements present in lower right are non-metals which are least reactive.

Hence, the correct answer is lower left.

You might be interested in

For each set of values, calculate the missing variable using the ideal gas law.

allsm [11]

Answer:

1. n = 0.174mol

2. T= 26.8K

3. P = 1.02atm

4. V = 126.88L

Explanation:

1. P= 2.61atm

V = 1.69L

T = 36.1 °C = 36.1 + 273= 309.1K

R = 0.082atm.L/mol /K

n =?

n = PV / RT = (2.61x1.69)/(0.082x309.1)

n = 0.174mol

2. P = 302 kPa = 302000Pa

101325Pa = 1atm

302000Pa = 302000/101325 = 2.98atm

V = 2382 mL = 2.382L

T =?

n = 3.23 mol

R = 0.082atm.L/mol /K

T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K

3. P =?

V = 0.0250 m³ = 25L

T = 288K

n = 1.08mol

R = 0.082atm.L/mol /K

P = nRT/V = (1.08x0.082x288)/25 = 1.02atm

4. P = 782 torr

760Torr = 1 atm

782 torr = 782/760 = 1.03atm

V =?

T = 303K

n = 5.26 mol

R = 0.082atm.L/mol /K

V = nRT/P

V = (5.26x0.082x303)/1.03 = 126.88L

8 0

3 years ago

If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr

xxMikexx [17]

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be $\frac{49}{40}=1.225$ g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

6 0

3 years ago

When might neutralization reactions be used in a laboratory setting?

Ipatiy [6.2K]

Neutralization reactions can be used in a laboratory setting in order to dispose of chemicals. When spills happens, for instance an acid is on the floor, you can use a base to neutralize the spill. Hope this answers the question. Have a nice day.

6 0

2 years ago

Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr

vredina [299]

Answer : The correct expression for equilibrium constant will be:

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$4A+2B\rightleftharpoons 8C$

The expression of $K_c$ will be,

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Therefore, the correct expression for equilibrium constant will be, $K_c=\frac{[C]^8}{[A]^4[B]^2}$

4 0

2 years ago

Try to make only one molecule of water in the simulation. Is it possible?

mel-nik [20]

Answer:

A single molecule of water has been isolated for the first time by trapping it in a fullerene cage. Water molecules are never found alone — they are always hydrogen-bonded to other molecules of water or polar compounds.

While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.

A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. The difference between atoms is expressed by atomic numbers.

Explanation:

3 0

3 years ago