Why does the reactivity of nonmetals increase moving left to right on the periodic table?

Chemistry

1 answer:

zhuklara [117]11 months ago

4 0

Answer: This is because the number of shell increases .

Explanation: On moving from left to write on the periodic table the reactivity of non metals increases because number of shells increases and the force with which the nucleus hold electrons decreases.

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9474 millimeters to centimeters using scientific notation showing work

puteri [66]

Answer:

9.474 x 10^2

Explanation:

ok. first you have to get the value in the required unit so 9474mm/(10mm/cm) = 947.4 so scientific notation states that the number must be raised to any power of an integer and the value of the number being raised must be less than than 10 and more than or equal to 1

so it must have one digit in front so.. 947.4 becomes 9.474 and because you move 2 places to the left, ur power is positive 2

and proof 10^2 is 100 so multiply 9.474 by 100 and u will get 947.4 cm which is also 9474 mm

8 0

3 years ago

What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

kari74 [83]

The heat of reaction (i.e. combustion) of butane ( $C_{4} H_{10}$ ) when reacted with oxygen ( $O_{2}$ )  is -2658 kJ/mol butane, and the chemical reaction is given by:

$C_{4} H_{10}$ + $\frac{13}{2} O_{2}$ ---> $4 CO_{2}$ + $5 H_{2}O$

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane

The mass of carbon dioxide ( $CO_{2}$ ) can be determined by multiplying the moles of butane ( $C_{4} H_{10}$ ) with the mole ratio of ( $CO_{2}$ ) produced to the ( $C_{4} H_{10}$ ) reacted, and then with the molar mass of ( $CO_{2}$ ), which is 44 grams/mole.

Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles $CO_{2}$ / 1 mole $C_{4} H_{10}$ ] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced
= 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.

4 0

2 years ago

Read 2 more answers

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$