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2 years ago

Which of these will water NOT dissolve?

Chemistry

1 answer:

kicyunya [14]2 years ago

7 0

<h2>.An oily cell membrane.</h2>

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Explain how water behaves in this reaction. Which definition of acids and bases would you apply?

mafiozo [28]

Water behaves as a base in this reaction.

The Bronsted-Lowry definition is applied, because the reaction involves the transfer of H+ from one reactant to the other.

A Bronsted-Lowry base is defined as a substance that accepts a proton.

Because water gains a proton to form H3O+ in this particular reaction, it acts as a base

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3 years ago

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Help me answer this please

Vlad1618 [11]

Its a scam dont press the link

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3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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4 years ago

During an experiment, a scientist places a heat lamp above a bowl of water and uses the lamp to heat up the water. How does heat

dlinn [17]

Answer:

Radiation

Explanation:

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The Earth produces a magnetic field in which the magnetic are most concentrated at the North and South

BlackZzzverrR [31]

Answer: I have no idea what kind of question this is, but I'm assuming it's a true or false question, and this is true. Although the north-bound pole is south-seeking, a compass would eventually draw you to the North Pole.

Explanation:

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