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3 years ago

What reaction illustrates the law of conservation of mass ?

1 answer:

8 0

The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases, equals the original mass of the charcoal and the oxygen when it first reacted. So the mass of the product equals the mass of the reactant

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How many electrons in an atom can have each of the following quantum number or sublevel designations?

lutik1710 [3]

Answer: a:6 electrons, b:10electrons, c: 2electrons

Explanation:

1.Principle quantum number(n)

n is the number of shell, it is a positive integer. The maximum number of electrons in a shell is 2(n2)

2.Azimuthal quantum number(l)

l is the number of subshell in the principal shell

The name of subshell are s, p, d, f

The number of nodes in each subshell is

s is l=0, p is l=1, d is l=2, f is l=3.

Each shell can have 2 x l + 1 sublevels, and each sublevel can accommodate maximum of 2 electrons and have two electrons.

A. n=2, l=1 then 2 x 1 + 1= 3 subshell, 3*2= 6 electrons in the p subshell

b. 3d, d is l = 2 then 2 x 2 + 1 = 5 sublevels, 5*2=10 electrons in the d subshell

c. 4s, s is l=0 then 2 x 0 + 1 = 1 sublevel, 1*2=2 electrons in the s subshell.

6 0

3 years ago

Why can some metals be extracted from compounds by heating with carbon and why can some not

pentagon [3]

Answer:

Some metals can be extracted from compounds by heating with carbon atom because they are less reactive than carbon and some metals cannot be extracted because they are more reactive than carbon atom.

Explanation: If the metal is less reactive than carbon atom so the carbon atoms make bond with oxide or other atom present with metal and the metal is free from that oxide or that element. But if the reactivity of metal is higher than carbon is unable to remove the oxide or element.

8 0

3 years ago

Find the volume of a gas at standard pressure if its volume at 1.9 atm is 80 ml?

kogti [31]

Answer:

1.5 × 10² mL

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 1.9 atm

Initial volume of the gas (V₁): 80 mL

Final pressure of the gas (P₂): 1.0 atm (standard pressure)

Final volume of the gas (V₂): ?

Step 2: Calculate the final volume of the gas

For an ideal gas, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1.9 atm × 80 mL/1.0 atm

V₂ = 1.5 × 10² mL

Since the pressure decreased, the volume of the gas increased.

6 0

3 years ago

An example of a colloid which is composed of a liquid dispersed in a gas would be

Aleksandr-060686 [28]

Any colloid consisting of a solid dispersed in a gas is called a smoke. A liquid dispersed in a gas is referred to as a fog. So the answer would be smoke

6 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago