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goldenfox [79]
2 years ago
12

What is neutralisation-acid and base?

Chemistry
1 answer:
lora16 [44]2 years ago
5 0
A neutralization reaction is when an <span>acid </span><span>and a </span>base react to form water and a salt and involves the combination of H+<span> ions and OH</span>-<span> ions to generate water .
Eg :-  </span><span>HCl + NaOH --> NaCl + H2O</span>
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The temperature at which a substance in the liquid phase transforms to the gaseous phase refers to the substance's _______. A. s
murzikaleks [220]
When a substance goes from being a liquid to a gas it evaporates, or boils away. Think of boiled eggs.
3 0
3 years ago
Read 2 more answers
Which of the following displays the correct change in enthalpy and best describes the reaction below? 2CsCl(aq) + Na2SO4(aq) 2Na
Bond [772]
We subtract the enthalpies of the reactants from that of the products:
2\Delta&#10; H(NaCl)+\Delta H(Cs_2 SO_4)-2\Delta H(CsCl)-\Delta H(Na_2 SO_4) \\ &#10;=2(-411)+(-1400) -2(-415)-(-1380) \\ = -12 kJ
Since this is < 0, this is an exothermic reaction.

3 0
3 years ago
Coke is an impure form of carbon that is often used in the in- dustrial production of metals from their oxides. If a sample of c
steposvetlana [31]

Answer:

The mass of coke needed to react completely with 1.0 ton of copper(II) oxide is 0.794 Ton.

Explanation:

2CuO+C\rightarrow 2Cu+CO_2

1 Ton = 907185 grams

Mass of copper oxide = 1.0 Ton = 907185 grams

Moles of copper oxide =\frac{907185 g}{79.55 g/mol}=11,403.95 moles

According to reaction, 2 moles of copper oxide reacts with 1 mole of carbon.

Then 11403.95 moles of copper oxide will react with:

\frac{1}{2}\times 11403.95 mol=5,701.98 mol of carbon

Mass of 5,701.98 moles of carbon:

5,701.98 mol\times 12 g/mol=68,423.75 g

Mass of coke = x

Mass of carbon = 68,423.75 g

Percentage of carbon in coke = 95%

95\%=\frac{68,423.75 g}{x}\times 100

x=720,250.09 g=0.794 Ton

The mass of coke needed to react completely with 1.0 ton of copper(II) oxide is 0.794 Ton.

3 0
3 years ago
It says draw a ring around all the objects that will be attracted to a magnet the one there are : a 2p coin,cobalt,an aluminium
Katen [24]

Answer: do

Explanation:

5 0
2 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
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