Question 1/5

The table shows the average distance between the molecules of a substance in the solid, liquid, and gaseous states.

Pavlova-9 [17]

Answer:

Y > X > Z

Explanation:

The intermoecular forces refer to forces that exist between molecules of a substance. They are the secondary bond forces that hold particles of a substance together in a particular state of matter.

The shorter the distance between molecules, the greater the magnitude of intermolecular force between the molecules.

The molecules of Y are at the shortest distance from each other hence they have the highest magnitude of intermolecular forces. Followed by X and lastly Z with the greatest distance between the largest intermolecular distance.

7 0

3 years ago

Which pair of elements would you expect to exhibit the greatest similarity in their physical and chemical properties? Sr, Te Ga,

Anvisha [2.4K]

Answer:

Explanation:

N and P will have similarity

Electronic configuration of N

2,5

1s²2s²p⁵

Electronic configuration of P

1s²2s²2p⁶3s²3p⁵

Electronic configuration bears similarity.

Both have 5 electrons in their outermost orbits . So they exhibit similarity in their properties. Both belong to the same 15 th group of periodic table.

Sr and Rb too bear much similarity . Both are strongly electro-positive metal , difference in their atomic no being only one.

5 0

3 years ago

(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum

ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0

2 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

25. 00 ml of a buffer solution contains 0. 500 m hclo and 0. 380 m naclo. if 50. 00 ml of water is added to the buffer, what are

geniusboy [140]

The new concentrations of $HClO$ and $NaClO$ are 0.25M and 19M

Calculation of number of moles of each component,

Molarity of $HClO$ = number of moles/volume in lit = 0. 500 M

Number of moles = molarity of $HClO$ × volume in lit = 0. 500 M× 0.025 L

Number of moles of $HClO$ = 0.0125 mole

Molarity of $NaClO$ = number of moles/volume in lit = 0. 38 M

Number of moles = molarity of $NaClO$ × volume in lit = 0. 38 M× 0.025 L

Number of moles of $NaClO$ = 0.95 mole

Calculation of new concentration at volume 50 ml ( 0.05L)

Molarity of $HClO$ = number of moles/volume in lit = 0.0125 mole/0.05L

Molarity of $HClO$ = 0.25M

Molarity of $NaClO$ = number of moles/volume in lit = 0.95mole/0.05L

Molarity of $NaClO$ = 19 M

learn about Molarity

brainly.com/question/8732513

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5 0

2 years ago