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katrin2010 [14]
3 years ago
11

Which of the following may suggest a catalyst has been used in a reaction, given the energy diagram for the same reaction withou

t a catalyst?

Chemistry
1 answer:
anzhelika [568]3 years ago
5 0
The answer I think would be C
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The fossil record provides evidence of past life as well as clues to the past climate, environment, and abiotic factors in an ar
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Arrange the bromine-containing compounds CaBr2, Br2, NBr3, CBr4 in order of increasing predicted boiling point.
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Explanation:

In an ionic compound, there will be strong force of attraction between the combining atoms due to the opposite charges present on them.

For example, CaBr_{2} is an ionic compound where calcium has a +2 charge and each bromine atom has a -1 charge.

Therefore, in order to break the bond we need to provide more heat. Hence, boiling point of calcium bromide will be the highest.

NBr_{3} is a covalent compound and as nitrogen is more electronegative in nature and also has a lone pair of electron hence, there will be a net dipole moment.

CBr_{4} is also a covalent compound. And, as bromine is more electronegative than carbon atom so, dipole moment is in the outwards direction. Hence, in CBr_{4} there will be zero dipole moment.

Therefore, its boiling is less than the boiling point of NBr_{3}.

Br_{2} is a covalent compound and there will be no dipole moment.

Thus, we can conclude that given bromine-containing compounds are placed in their increasing boiling point order as follows.

    Br_{2} < CBr_{4} < NBr_{3} < CaBr_{2}

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4 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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Answer:

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