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3 years ago

The reducing agent in the reaction described in Fe + 2HCl → FeCl2 + H2 is A. H2. B. Fe. C. HCl. D. FeCl2.

Chemistry

2 answers:

Fed [463]3 years ago

7 0

The reducing agent is b). Fe

dalvyx [7]3 years ago

5 0

the answer is Fe i took the test

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Consider the reaction. CaCl2(aq)+K2CO3(aq)⟶CaCO3+2KCl. Identify the precipitate, or lack thereof, for the reaction. (A) KCl (B)

Natali5045456 [20]

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

Precipitate is defined as insoluble solid substance that emerges when two different aqueous solutions are mixed together. It usually settles down at the bottom of the solution after sometime.

For the given chemical equation:

$CaCl_2(aq.)+K_2CO_3(aq.)\rightarrow CaCO_3(s)+2KCl(aq.)$

The products formed in the reaction are calcium carbonate and potassium chloride. Out of the two products, one of them is insoluble which is calcium carbonate. Thus, it is considered as a precipitate.

Hence, the correct answer is Option B.

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4 years ago

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m

Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

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

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

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, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 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 = $87ml+87ml=174ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g$

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 = 174 g

$T_{final}$ = final temperature of water = 317.4 K

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

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

$q=174g\times 4.18J/g^oC\times (317.4-298)K$

$q=14110.008J=14.11KJ$

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

$\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole$

Therefore, the enthalpy of neutralization is, -101.37 KJ

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