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2 years ago

What changes if the degree (amount) of intermolecular bonds of a substance are changed?

Chemistry

2 answers:

elixir [45]2 years ago

7 0

<h2>Answer:</h2>

The correct answers are;

volume of a substance

density of a substance

The volume and the density of the substance depend on inter molecular forces of the substance. So, when we will change the degree of inter molecular forces, whether decrease it or increase it, it will affect the volume and density of a substance.

Nuetrik [128]2 years ago

6 0

<u>Answer:</u> The correct statements are volume of a substance and density of a substance.

<u>Explanation:</u>

Degree of intermolecular bonds is defined as the forces by which two particles are bonded in a substance. If the intermolecular forces are strong, it means that it will occupy less space or volume and hence, the density of that substance will be more.

If the intermolecular forces are weak, it means that it will occupy less space and density of the substance will be more.

Relationship between volume and density of a substance is given by the relation:

$Density=\frac{Mass}{Volume}$

Density and volume has inverse relationship. If volume increases, density will decrease and vice-versa.

Hence, the correct statements are volume of a substance and density of a substance.

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A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

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2 years ago

Explain how manipulation of light waves can cause reflection, refraction, diffusion, and absorption; and describe how different

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Answer:is A

Explanation:

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2 years ago

Four different solutions (I, II, III, and IV) are labeled on the pH scale below. A p H scale is shown, fading from red at the le

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Answer: It has the highest number of hydroxide ions

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2 years ago

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Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle

Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

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2 years ago

The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia

tiny-mole [99]

Answer:

The percent yield of this reaction is 70%

Explanation:

The reaction is: N₂ + 3H₂ → 2NH₃

We only have the mass of H₂, so we assume that N₂ is in excess

We convert the mass to moles, to work with the reaction:

450 g . 1mol / 2 g = 225 moles

Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia

Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles

This is the 100% yield reaction → We convert the moles of NH₃ to mass

150 mol . 17g /1mol = 2550 g

Percent yield = (Produced yield/Theoretical yield) .100

Percent yield = (1575g/2550g) . 100 = 70%

7 0

2 years ago