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Oksanka [162]
3 years ago
12

A sample of 25mls of HCL was mixed with 25mls of KOH of the same concentration in a calorimeter. As a result the temperature ros

e from 25'c to 26.6'c. Given that the enthalpy is -56.02kj/mol for the reaction and the specific heat of the dilute aqueous solution is approximately equal to that of water (4.184j/g'C). Determine the molarity of the original HCL solution.
Chemistry
1 answer:
xxTIMURxx [149]3 years ago
4 0
Ooo that cool good thing I learn from this app
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Draw the major organic product(s) for the reaction. The starting material is a benzene ring with one substituent. The substituen
Mice21 [21]

Answer:

See Explanation

Explanation:

In electrophilic aromatic substitution, the benzene ring undergoes substitution when it is reacted with suitable electrophiles.

The products of electrophilic aromatic substitution depends on the substituents already present on the benzene ring. Some substituents activate the ring towards electrophilic substitution and direct the incoming electrophile to the ortho and para positions on the ring while some substituents deactivate the benzene ring towards electrophilic substitution and direct the incoming electrophlle to the meta position on the ring.

The amide substituent is moderately activating and is an ortho, para director hence the products shown in the mage attached to this answer.

3 0
2 years ago
Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

  • Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

  • Divide by the smallest mol(Mn=0.0115)

Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
3 years ago
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