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3 years ago

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

Answer and Explanation:

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

Naya [18.7K]

Answer:

Physical Properties

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.

Chemical Properties

1. Metals are good conductorsof heat and electricity but most non - metals are insulators.

2. Metals, when exposed to water atmospheric oxygen tend to rust but non - metals lack this chemical property

7 0

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

erastova [34]

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_________

posledela

I think the answer to this is A

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3 years ago

Read 2 more answers

When studying seismological data scientists use data from...

Dovator [93]

I do not understand science but if u ask me I would have no clue do u get what I mean

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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

Molar mass of $N_2$ = 28 g/mole

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