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3 years ago

10.

Chemistry

1 answer:

Andru [333]3 years ago

7 0

Answer: 1.8 moles Fe and 2.7 moles $CO_2$ are produced.

Explanation:

The balanced chemical reaction is:

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

According to stoichiometry :

3 moles of $CO$ require 1 mole of $Fe_2O_3$

Thus 2.7 moles of $CO$ will require= $\frac{1}{3}\times 2.7=0.9moles$ of $Fe_2O_3$

Thus $CO$ is the limiting reagent as it limits the formation of product and $Fe_2O_3$ is the excess reagent.

As 3 moles of $CO$ give = 2 moles of $Fe$ and 3 moles of $CO_2$

Thus 2.7 moles of $CO$ will give = $\frac{2}{3}\times 2.7=1.8moles$ of $Fe$ and $\frac{3}{3}\times 2.7=2.7moles$ of $CO_2$

Thus 1.8 moles Fe and 2.7 moles $CO_2$ are produced.

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Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,

VARVARA [1.3K]

Answer:

If it receives energy faster than it can radiate/convect/conduct it away, its temperature will increase. It does not matter.

Explanation:

6 0

3 years ago

Read 2 more answers

A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go

4vir4ik [10]

Answer : The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of $NH_4Cl$ = 26.5 g

Mass of $NaOH$ = 10 g

Molar mass of $NH_4Cl$ = 53.5 g/mole

Molar mass of $NaOH$ = 40 g/mole

First we have to calculate the moles of $NH_4Cl$ and $NaOH$ .

$\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles$

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NH_4Cl+NaOH\rightarrow NH_4OH+NaCl$

From the balanced reaction we conclude that

As, 1 mole of $NaOH$ react with 1 mole of $NH_4Cl$

So, 0.25 moles of $NaOH$ react with 0.25 moles of $NH_4Cl$

From this we conclude that, $NH_4Cl$ is an excess reagent because the given moles are greater than the required moles and $NaOH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of $NH_4Cl$ .

$\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl$

$\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g$

Therefore, the amount of reactant left in excess is 13.1075 grams.

5 0

3 years ago

An empty Erlenmeyer flask weighs 241.3 g. When filled with water (d = 1.00 g/cm³), the flask and its contents weigh 489.1 g.

Viefleur [7K]

The answer is- Mass of flask = 608.04 g

Density (d) is defined as mass of substance divided by its volume. Thus, if mass and density are known, then Volume can be determined.

What is the formula of mass in terms of density?

Let mass of substance be 'm' and volume be 'V'. Thus, as per the definition of density, it is expressed as-

$d = \frac{m}{V}$

Thus, mass can be expressed as-

$m = d * V$

Now, mass of empty flask = 241.3 g and the mass of (flask + Water) = 489.1 g.

Thus, mass of water = $489.1\ g- 241.3\ g = 247.8\ g$ .

Thus, mass of water = 247.8 g and density of water = $1.00\ g/cm^3$ . Its volume is calculated as-

$V = \frac{247.8\ g}{1.00\ g/cm^3}=247.8\ cm^3$

Volume of flask = 247.8 $cm^3$ .
Then when this flask is filled with chloroform (d = $1.48 g/cm^3$ ), the mass of chloroform is-

$m = d * V = (1.48\ g/cm^3) * (247.8\ cm^3)= 366.74\ g$

Hence, the mass of flask becomes = $241.3\ g + 366.74\ g = 608.04\ g$

To learn more about Density and mass, visit:

brainly.com/question/952755

#SPJ4

4 0

1 year ago

[1] b) In a local election there were 8000 registered voters. Of these 3200 voted. What was the percentage that voted? [1] <

Alexandra [31]

Answer:

40% of registered voters voted

Explanation:

3200 out of 8000,

or :

3200 / 8000

= 0.4

0.4 x 100 = 40

40% of registered voters voted

4 0

2 years ago

Determine each type of reaction. 2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l ) 2CX2HX2(g)+5OX2(g)⟶4COX2(g)+2HX2O

Valentin [98]

Answer:

2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction

N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction

C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction

2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction

C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction

Explanation:

We can determine the type of reaction by considering the reactants and products.

Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.

A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.

A combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form methanol.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).

A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.

4 0

4 years ago