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pishuonlain [190]
3 years ago
13

When using evaporation/boiling, it is

Chemistry
1 answer:
densk [106]3 years ago
3 0
The answer would be b) GAS
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Lead is malleable, so it can be pounded into flat sheets without breaking. How does the bonding within lead help to explain this
damaskus [11]

Answer:

correct answer would be c

8 0
2 years ago
Determine the mass in grams of each element.
jeka57 [31]

1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g

2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g

3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g

4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g

<h3>1. Determination of the mass of 1.33×10²² mole of Sb</h3>
  • Mole of Sb = 1.33×10²² mole
  • Molar mass of Sb = 122 g/mol
  • Mass of Sb =?

Mass = mole × molar mass

Mass of Sb = 1.33×10²² × 122

Mass of Sb = 1.62×10²⁴ g

<h3>2. Determination of the mass of 4.75×10¹⁴ mole of Pt</h3>
  • Mole of Pt = 4.75×10¹⁴ mole
  • Molar mass of Pt = 122 g/mol
  • Mass of Pt =?

Mass = mole × molar mass

Mass of Pt = 4.75×10¹⁴ × 195

Mass of Pt = 9.26×10¹⁶ g

<h3>3. Determination of the mass of 1.22×10²³ mole of Ag</h3>
  • Mole of Ag = 1.22×10²³ mole
  • Molar mass of Ag = 108 g/mol
  • Mass of Ag =?

Mass = mole × molar mass

Mass of Ag = 1.22×10²³ × 108

Mass of Ag = 1.32×10²⁵ g

<h3>4. Determination of the mass of 9.85×10²⁴ mole of Cr</h3>
  • Mole of Cr = 9.85×10²⁴ mole
  • Molar mass of Cr = 52 g/mol
  • Mass of Cr =?

Mass = mole × molar mass

Mass of Cr = 9.85×10²⁴ × 52

Mass of Cr = 5.12×10²⁶ g

Learn more about mole:

brainly.com/question/13314627

7 0
2 years ago
Describe the oxygen/carbon dioxide cycle using plants and animals.
Vikentia [17]

Answer:

The process of photosynthesis in plants releases oxygen into the atmosphere. Respiration by plants and animals, as they use the energy stored in food, and the process of decomposition of dead organisms, releases carbon dioxide into the atmosphere. All three work together to maintain the carbon dioxide-oxygen cycle.

5 0
3 years ago
Entropy of a system decreases with decrease in Temperature T/F
Svet_ta [14]

Answer: The given statement is true.

Explanation:

Entropy means the measure of randomness present in a substance. That is, an increase in temperature will lead cause more motion in the particles of a substance more will be their kinetic energy.

As a result, there will occur more collisions due to which randomness of molecules will increase. Hence, there will be increase in entropy.

So, when we decrease the temperature then there will be decrease in motion of particles. As a result, lesser number of collisions will take place between them. Hence, degree of randomness will also decrease.

Thus, we can conclude the statement entropy of a system decreases with decrease in temperature, is true.

3 0
3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

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

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

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

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
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