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jeka94
3 years ago
15

What does ozone do to make life on earth possible?

Chemistry
1 answer:
forsale [732]3 years ago
6 0
Answer - Ozone layer helps keep the harmful ultra violet rays hitting and keeping the atmospheric pressure in.

Reasoning - Without the ozone layer the whole planet would have a loss of water and especially people burning very easily from the sun. <span />
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Which is one way to test whether an unknown solution is acidic or basic
yarga [219]

if the ph is above 7 then its acidic

if its below 7 than its going to be basic.

~serenity bella

7 0
3 years ago
Read 2 more answers
Two alcohols, isopropyl alcohol and propyl alcohol, have the same molecular formula, C3H8O. A solution of the two that is two-th
iris [78.8K]

Answer:

1.

Since both components of these solutions have the same molar mass, mole fractions would be the same as mass fractions.

0.110 atm = (2/3)(Pi) + (1/3)(Pn) [1]

0.089 atm = (1/3)(Pi) + (2/3)(Pn) [2]

2*[1] - [2]:

(2)(0.110) - 0.089 atm = Pi

Pi = 0.131 atm

2*[2] - [1]:

(2)(0.089) - 0.110 atm = Pn

Pn = 0.068 atm

2.

The hydroxyl (-OH) group on the end of a longer 1-propanol molecule makes it more polar than IPA. It follows that the intermolecular forces between 1-propanol are stronger than those of IPA and thus the vapor pressure of 1-propanol should be lower than IPA.

Explanation:

5 0
4 years ago
Are photosynthesis and cellular respiration part of the land-based carbon cycle of water-based carbon cycle
Dmitrij [34]
It would be the water based carbon cycle
3 0
3 years ago
For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
  • brainly.com/question/19671384
5 0
3 years ago
How much heat is required to change the temperature of a 15 g aluminum can with 100 g of water from 24.5°C to 55°C?
Assoli18 [71]
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium 
Heat gained by water = 100 × 4.186 × 30.5 
                                   = 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5 
                                          = 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
3 0
3 years ago
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