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Galina-37 [17]
3 years ago
5

How have humans contributed to an increase in carbon dioxide levels in the past 100 years?

Chemistry
1 answer:
deff fn [24]3 years ago
8 0

Answer:

Carbon dioxide (CO₂) is an important green house gas. A small increase in the concentration of this gas can lead to the global rise in the surface temperature. Over the last 100 years, this gas has increased significantly. One of the main cause behind this the anthropogenic activities.

After the setup of the industrial revolution, there has been a large increase in this CO₂ concentration, as there are numerous large factories and indutries that releases this harmful gases. In addition to this, the burning up of fossil fuel also is responsible for the emission of large amount of carbon into the atmosphere.

Deforestation also is another cause, because trees are no longer there in vast quantities which takes up a large number of atmospheric carbon and in return liberates plenty of oxygen.

Thus, the activities done by man over the last century has led to the increase in the concentration of CO₂ into the atmosphere.

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What is the boiling point of cyclohexane at 620 mm hg?
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Student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combin
nirvana33 [79]

Answer:-  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}

Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.

To calculate the heat absorbed or released we use the formula:

q=ms\Delta T

q = heat absorbed or released

m = mass of solution

s = specific heat capacity

and \Delta T = change in temperature

mass of solution = mass of solute + mass of solvent

mass of solution = 5.00 g + 100.0 g = 105.0 g

(note:- density of pure water is 1 g per mL so the mass is same as its volume)

\Delta T = 33.0 - 23.0 = 10.0 degree C

s = 4.18\frac{J}{g.^0C}

Let's plug in the values in the formula and calculate q.

q = 105.0g(4.18\frac{J}{g.^0C})(10.0^0C)

q = 4389 J

To calculate the enthalpy of solution that is \Delta H_s_o_l_n we convert q to kJ and divide by the moles of solute.

moles of solute = 5.00g(\frac{1mol}{42.39g})

= 0.118 moles

q = 4389J(\frac{1kJ}{1000J}) = 4.389 kJ

\Delta H_s_o_l_n=\frac{4.389kJ}{0.118mol}

\Delta H_s_o_l_n = 37.2\frac{kJ}{mol}

Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.

So,  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}


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3 years ago
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