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2 years ago

8

Which pair of elements would you expect to exhibit the greatest similarity in their physical and chemical properties? o,s si, p?

1 answer:

aliya0001 [1]2 years ago

8 0

I believe the appropriate pair is O and S, Periodic table represents arrangements of elements ordered by their atomic number,physical and chemical properties. Both sulfur and oxygen are in the same group of the periodic table (group 6), they have a valency of 2, This means they have a greater similarity in chemical and physical properties.

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A sample of gas consists of 4.0 moles, and exists at stp. What is the volume of the gas?

faust18 [17]

Answer:

The volume of the gas is 89.60

Explanation:

When converting from moles to volume, you would multiply by 22.4 because 1 mole equals 22.4 liters at STP.

So...

4.0 moles * 22.4 liters =

89.60 liters

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2 years ago

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A van moves with a constant speed

Kitty [74]

Answer:

1.5 hours or 90 minutes

Explanation:

Velocity = d/t

V = 64 mi / hour or 96 mi / 9 min OR 16 mi / 15 min

D = 96 miles

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T = 96 mi / 64 mi / hour

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2 years ago

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Daniel [21]

one mole of P weights about 31 grams

in one mole there are 6.022*10^23 atoms

we use the rule of threes

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2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

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

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

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Therefore, the value of equilibrium constant for this reaction is, 1.1

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2 years ago

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Usimov [2.4K]

Answer:

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Explanation:

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3 years ago