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3 years ago

When a substance is in the liquid state, how are the particles of that substance behaving

Chemistry

2 answers:

iVinArrow [24]3 years ago

4 0

Answer:

In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles. With an increase in temperature, the particles move faster as they gain kinetic energy, resulting in increased collision rates and an increased rate of diffusion.

Explanation:

solong [7]3 years ago

3 0

In short, the particles are more loose and moving

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What would be the advantage of using a net ionic equation to represent a redox reaction

wolverine [178]

The net ionic equation is shorter to use and already leaves out the electrons that transferred from the reducing agent to the oxidizing agent. Also, in some occasions the aqueous ions H+ and (or) OH- ions that help balance the net ionic charge are no longer shown in the net ionic equation.

5 0

4 years ago

The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air

ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ = ΔT + T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0

3 years ago

At the freezing point, the particles in an object have no kinetic energy. true or false?

Sliva [168]

False because particles stop moving or study slow down.

5 0

3 years ago

Read 2 more answers

An alkyne with the molecular formula C5H8 was reduced with H2 and Lindlar's catalyst. Hydroboration-oxidation of the resulting a

irinina [24]

Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.

The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.

When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.

Learn more: brainly.com/question/2510654

6 0

3 years ago

Select the correct answer.

muminat

Answer:- B. 4.65 g.

Solution:- The given balanced equation is:

$2AgNO_3(aq)+Na_2S(aq)\rightarrow Ag_2S(s)+2NaNO_3(aq)$

It asks to calculate the mass of silver sulfide formed by when 0.0150 liters of 2.50 M of silver nitrate are used.

Moles of silver nitrate are calculated on multiplying it's liters by its molarity and then on multiplying by mol ratio, the moles of silver sulfide are calculated. These moles are multiplied by the molar mass to convert to the grams.

Molar mass of $Ag_2S$ = 2(107.87)+32.06 = 247.8 g per mol

The dimensional set up for the complete problem is:

$0.0150L(\frac{2.50molAgNO_3}{1L})(\frac{1molAg_2S}{2molAgNO_3})(\frac{247.8gAg_2S}{1molAg_2S})$

= $4.65gAg_2S$

So, the correct choice is B. 4.65 g.

7 0

3 years ago