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kow [346]
2 years ago
15

What was the primary cause of the increase in oxygen content between 400 and 300 million years ago

Chemistry
1 answer:
Gelneren [198K]2 years ago
4 0

Answer:

What was the primary cause of the increase in oxygen content between 400 and 300 million years ago

Explanation:

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Give the name of the products formed
attashe74 [19]

Answer:

With Br2 - Bromobenzene

With Cl2 - Chlorobenzene

With HNO3- Nitrobenzene

With H2SO4 - Benzenesulphonic acid

With HCOCl - Benzoyl chloride

With 1-chloro-2,2-dimethylpropane - 2,2dimethyl-1-phenyl propane

Explanation:

The common thread joining all these reactions is that they are all electrophillic reactions. They are so called because the attacking agents in each reagent is an electrophile. Electrophiles are species that have electron deficient centers and are known to attack molecules that are high in electron density at regions of high electron density.

The benzene molecule has rich electron density. Any substituents that donates electrons to the ring improves the likelihood that benzene will undergo electrophillic substitution reactions while electron withdrawing substituents decrease the likelihood that benzene will undergo electrophillic substitution reactions.

The names of the compounds formed when benzene undergoes electrophillic reaction with the attacking agents listed in the question are displayed in the answer section.

7 0
2 years ago
How would you describe a food chain to someone who never thought about that idea before
umka21 [38]
A food chain is a sequence going from the producers on the bottom to the consumers to the top that shows what consumer eats what
6 0
2 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
2 years ago
Who knows how to do this please help
Fofino [41]

Answer: You forgot to zero the balance

Explanation:

5 0
2 years ago
A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150
Usimov [2.4K]

Answer:

442.3 mL

Explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

M=\frac{Moles of substance X}{Volume of the solution}

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

11.27 g AgNO_{3} *\frac{1 mole AgNO_{3}}{169.87 g AgNO_{3}} = 0.06634 moles AgNO_{3}

Then you can solve for the volume of the solution:

Volume of the solution=\frac{Moles of AgNO_{3}}{M} =\frac{0.06634 mol AgNO_{3}}{0.150 M} =0.4423 L = 442.3 mL

Hope it helps!

3 0
3 years ago
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