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Paha777 [63]
3 years ago
13

In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury

Chemistry
1 answer:
Nuetrik [128]3 years ago
5 0

<u>Answer and Explanation:</u>

Mercury combines with sulfur as follows -

Hg + S = HgS

Hg = 200,59

S = 32,066 Therefore 1.58 g of Hg will react with -

1.58 multiply with 32,066 divide by 200,96 of sulfur.

= 0.25211 g S

This will form 1.58 + 0.25211 g HgS  = 1.83211 g HgS

The amount of S remaining = 1.10 - 0.25211  = 0.84789 g

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State two differences between metals and non metals with respect to their
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Answer:

<u>Physical Properties</u>

1. Metals are shiny but most non - metals lack this property.

2. Metals are able to deform under compression (malleable) but most non - metals lack this property.

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3 years ago
How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s
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Answer : The grams of carbon monoxide needed are 148.89 g

Solution : Given,

Mass of iron, Fe = 198.5 g

Molar mass of iron, Fe = 56 g/mole

Molar mass of carbon monoxide, CO = 28 g/mole

First we have to calculate the moles of iron, Fe.

\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles

The balanced chemical reaction is,

Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)

From the balanced reaction, we conclude that

2 moles of iron produces from the 3 moles of carbon monoxide

3.545 moles of iron produces from the \frac{3}{2}\times 3.545=5.3175 moles of carbon monoxide

Now we have to calculate the mass of carbon monoxide, CO.

\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}

\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g

Therefore, the grams of carbon monoxide needed are 148.89 g

7 0
2 years ago
The area in which certain types of plants or animals can be found living in close proximity to each other is called a_________
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3 years ago
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

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Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
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