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1 year ago

15

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

1 answer:

zhuklara [117]1 year 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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statement 1,mass is conserved durning the reaction. statement 2 sum of mass and energy is conserved durning the reaction. which

maksim [4K]

Answer:

Explanation:

None of the statement is true for both chemical and nuclear reactions. In chemical reactions, mass is always conserved and the type of atoms are also conserved.

6 0

3 years ago

How many moles are in 20 grams of argon

Sladkaya [172]

Answer:

There are 0.5 mole in 20g of argon.

Explanation:

40 g of argon = 1mole

Then 20g of argon is,

→ 1/40 × 20

→ 0.5 mole

5 0

2 years ago

Which of the following Group 1A elements is the most reactive?

Alisiya [41]

A. Cesium because reactivity of alkali metals increases from the top to the bottom of the group.

7 0

3 years ago

Read 2 more answers

All of the following are examples of matter except

Alexxandr [17]

A. Heat because it does not take up space.

3 0

3 years ago

In an experiment, a 0.4351 g sample of benzil (C14H10O2) is burned completely in a bomb calorimeter. The calorimeter is surround

andre [41]

Answer:

The enthalpy change during the reaction is -7020.09 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $994.1 J/^oC$

$T_{final}$ = final temperature = $27.60^oC$

$T_{initial}$ = initial temperature = $25.10^oC$

Now put all the given values in the above formula, we get:

$q=994.1 J/^oC\times (27.60-25.10)^oC$

$q= 2,485.25 J$

The heat gained by water present in calorimeter. = q'

$q'=mc\times (T_{final}-T_{initial})$

where,

q' = heat gained = ?

m = mass of water = $1.153\times 10^3 g$

c' = specific heat of water = $4.184 J/^oC$

$T_{final}$ = final temperature = $27.60^oC$

$T_{initial}$ = initial temperature = $25.10^oC$

$q'=1.153\times 10^3 g\times 4.184 J/^oC\times (27.60-25.10)^oC$

q ' = 12,060.38 J

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J

Q = -14.54563 kJ

n = number of moles fructose = $\frac{\text{Mass of benzil}}{\text{Molar mass of benzil}}=\frac{0.4351 g}{210 g/mol}=0.002072 mole$

$\Delta H=-\frac{-14.54563 kJ}{0.002072 mole}=-7020.09 kJ/mole$

Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.

3 0

3 years ago