What molecule experience hydrogen bonding

List 2 extensive properties and 3 intensive properties of a 5.0g, 1 cm3 cube of silver

jeka57 [31]

Properties of matter can be broadly classified into two categories:

Physical properties which usually involve a change in the state of matter and Chemical properties which involve a change in the chemical composition of matter.

Now, physical properties can be further classified as:

Extensive: these depend on the amount of the substance, eg: mass, volume

Intensive: these do not depend on the amount of the substance eg: density, color, melting point, boiling point

Here we are given a 5.0 g and 1 cm3 silver cube :

Therefore:

Extensive properties are-

1) Mass of silver = 5.0 g

2) Volume of silver = 1 cm3

Intensive properties are:

1) Density of silver = mass/volume = 5.0 g/ 1 cm3 = 5.0 g/cm3

2) Melting point of silver = 962 C

3) Color = white/gray

3 0

3 years ago

What is the noble gas electron configuration of aluminum (Al)? A. [Ne] 3s2 3p1 B. [Ne] 2s2 2p1 C. [Ar] 3s2 3p1 D. [Ar] 1s2 2s2 2

Gekata [30.6K]

[Ne] 3s2 3p1 So its A

6 0

3 years ago

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How can you measure the volume of an irregular shaped solid?

Archy [21]

If the object has an irregular shape, the volume can be measured using a displacement can. The displacement can is filled with water above a narrow spout and allowed to drain until the water is level with the tap. As the irregular object is lowered into the displacement can, the water level rises.

7 0

3 years ago

At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO

erastovalidia [21]

Answer : The concentration of $NO$ at equilibrium is 0.9332 M

Solution : Given,

Concentration of $N_2$ and $O_2$ at equilibrium = 0.200 M

Concentration of $N_2$ and $O_2$ at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}$

$K_c=6.25$

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}$

By solving the term x, we get

$x=2.6\text{ and }-0.0666$

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of $NO$ at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of $NO$ at equilibrium is 0.9332 M

5 0

3 years ago

If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed.

olga nikolaevna [1]

Super saturated solution is formed.

<u>Explanation:</u>

Solubility is the property of any substance's capacity, that is the solute of the substance is dissolved in the given solvent to form the solution. We have three different types of solution, unsaturated, saturated and supersaturated solution.

Unsaturated solution is a solution with lesser amount of solute than its solubility at equilibrium.

Saturated solution is a solution with the maximum solute dissolved in the solvent.

Super saturated solution is a solution with more solute than it is required.

The solubility of KI at 30°C is 153 g / 100 ml. Here 180 g of KI in 100 ml of water at 30°C is given, which has more solute than required, so it is super saturated solution.

6 0

3 years ago

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