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2 years ago

Https://ny.pbslearningmedia.org/resource/chemical-engineering-your-future/chemical-engineering-video/

1 answer:

8 0

Both disciplines deal with much the same things and there is much overlap, but the basic differences are novelty and scale. It is sometimes said that a chemist wears a lab coat; a chemical engineer wears a hard hat.

Explanation:

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

2 years ago

Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC

Anastaziya [24]

Answer:

There are 1.287 grams of acetylene collected

Explanation:

Total gas pressure = 909 mmHg

Vapor pressure of water = 20.7 mmHg

Pressure of acetylene = 909 mmHg - 20.7 mmHg = 888.3 mmHg

1mmHg = 1 torr

22 ° C + 273.15 = 295.15 Kelvin

Ideal gas law ⇒ pV = nRT

⇒ with p = pressure of the gas in atm

⇒ with V = volume of the gas in L

⇒ with n = amount of substance of gas ( in moles)

⇒ with R = gas constant, equal to the product of the Boltzmann constant and the Avogadro constant (62.36 L * Torr *K^−1 *mol^−1)

⇒ with T = absolute temperature of the gas (in Kelvin)

888.3 torr * 1.024 L = n * 62.36 L * Torr *K^−1 *mol^−1 * 295.15 K

n = 0.04942 moles of C2H2

Mass of C2H2 = 0.04942 moles x 26.04 g/mole = 1.287 g

There are 1.287 grams of acetylene collected

6 0

2 years ago

6 At STP, which element is a good conductor of electricity?(1) chlorine (3) silver

alexgriva [62]

For an object to conduct electricity it should have free or delocalised electrons that are free to pass the charge and hence take part in conducting electricity.
From the given choices
Chlorine is a halogen existing as a diatomic gas. Iodine too is a halogen and 2 Iodine atoms held together by covalent bond. Cl - Cl bonds and I-I bonds are covalent bonds. the outer electrons of Cl and I take part in covalent bonds therefore they are fixed and not free to move about. therefore no free electrons to conduct electricity.
Sulfur is a solid that too is held together by covalent bonds so it does not have free electrons to conduct electricity.

Silver is a metal and a general property of metals are their ability to conduct electricity.
metal structure are metal ions tightly packed together. when the metal atoms are tightly packed their valence electrons are removed and delocalised. Positively charged metal ions are embedded in a sea of delocalised electrons.
therefore there are delocalised electrons that can conduct electricity

answer is 3) silver

3 0

2 years ago

Read 2 more answers

This substance consists of two or more different elements chemically combined

kow [346]

The substance would be referred to as a compound substance.

8 0

2 years ago

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.0 g of lead(II) oxide. Calculate the percent yi

user100 [1]

Let MM(x) be the molar mass of x.

MM(Pb) : MM(PbO)
=207.21 : 223.20 = 451.4 g : x g

cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g

Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)

4 0

3 years ago