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My name is Ann [436]
3 years ago
15

Can you give examples of carbon in our living environment?

Chemistry
2 answers:
sveticcg [70]3 years ago
5 0
Wastes
Fuel
Smoke
Gas
Toxic chemicals
REY [17]3 years ago
4 0
Our own bodies contain carbon - coal as well and plastics
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Help me get this done today please no wrong answer
myrzilka [38]

Answer:

d.

Explanation:

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5 0
3 years ago
What is the mass of 2.0 moles of CO2
igor_vitrenko [27]
1 mol of CO2 is 44.01g/mol
So multiply that by 2 to get 2 mol of CO2, which is 88.02g
6 0
4 years ago
An experiment was conducted to see which metal pot would be better for cooking food quickly. The table below shows how quickly s
LenaWriter [7]

Answer:

The term conclusion best illustrates the given statement.

Explanation:

Conclusion refers to a decision or judgment that can be acquired by reasoning. From the observation mentioned in the given table:  

1. The steel, aluminum, and bronze took 27, 45 and 80 seconds to dissolve small section of wax, this signifies that more time will be needed by them to transfer heat to the food while cooking. Thus, food will take more time to cook in them.  

2. Copper consumed less time, that is, 15 seconds to melt the section of wax than the other metal pots, which signifies that it is a good conductor of heat than the others. Thus, the copper pot would be good for cooking food briskly.  

After evaluating the results and observations in an experiment, the conclusion was made that pots made of copper are best for cooking foods.  

4 0
3 years ago
What is a substances heat capacity?
Phantasy [73]

Also called thermal capacity, is when the amount of heat required to raise the temp. of one mole or one gram of a substance by one degree Celsius without change of phase.


Hope this helps :)

6 0
3 years ago
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains
erastova [34]

Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0
3 years ago
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