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Alecsey [184]
2 years ago
9

Iron(II) sulfide reacts with hydrochloric acid according to the reaction:

Chemistry
2 answers:
Inessa [10]2 years ago
8 0
The balanced chemical reaction would be:

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g) 

We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:

Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl

0.646 - 0.48 = 0.166 mol HCl left
zysi [14]2 years ago
4 0

Answer: The amount (in moles) of the excess reactant left is, 0.166 mol

Explanation : Given,

Moles of FeS = 0.240 m ol

Moles of HCl = 0.646 mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)

From the balanced reaction we conclude that

As, 1 mole of FeS react with 2 mole of HCl

So, 0.240 moles of FeS react with 0.240\times 2=0.48 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and FeS is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of HCl

Remaining moles of HCl = 0.646 - 0.48 = 0.166 mol

Thus, the amount (in moles) of the excess reactant left is, 0.166 mol

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You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially
nika2105 [10]

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is 0.0944J/g^oC

Explanation :

<u>For part A :</u>

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the metal

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 51.5J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 50.0 g

\Delta T = change in temperature = (T_{final}-T_{initial})=23.2-22.0=1.2^oC

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

q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]

q=312.48J

Thus, the heat released by the metal is -312.48 J

<u>For part B :</u>

q=m\times c\times \Delta T

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

\Delta T = change in temperature = (T_{final}-T_{initial})=23.2-100=76.8^oC

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

-312.48J=43.1g\times c\times 76.8^oC

c=0.0944J/g^oC

Thus, the specific heat of the metal is 0.0944J/g^oC

8 0
3 years ago
If you wanted to buy an aquarium with only benthos in it which of the following would you buy
Tju [1.3M]

Answer:

"Organisms in this tank are attached to the bottom, but use the sun to make food."

Explanation:

<em>"The principal food sources for the benthos are plankton and organic debris from land. In shallow water, larger algae are important, and, where light reaches the bottom, benthic photosynthesizing diatoms are also a significant food source."</em>

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3 years ago
Identify oxidation.
nevsk [136]

Oxidation is "Increase in oxidation number" as well as loss of electrons.

A rise in oxidation number results from the loss of negative electrons, whereas a reduction in oxidation number results from the gain of electrons. As a consequence, the oxidized element or ion experiences a rise in oxidation number.

As a result of losing electrons in the process, a reactant oxidizes. When a reactant obtains electrons during a reaction, reduction takes place. This frequently happens when acid and metals react.

Therefore, Oxidation is "Increase in oxidation number" as well as loss of electrons.

Hence, the correct answer will be option (e)

To know more about Oxidation

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8 months ago
How does this splitting wedge make work easier?
nikitadnepr [17]
The ansewr is A i would guess

7 0
3 years ago
How many grams of glucose are in 11.5 moles?
qaws [65]

There are 2071.4662 grams of glucose in 11.5 moles.

Per 1 mole there are 180.15588 grams of glucose. 180.5588 x 11.5 =2076.4262

5 0
3 years ago
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