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Usimov [2.4K]
1 year ago
9

What are the states of the following reaction C3H7Br +H2O --> C3H7OH +HBr

Chemistry
1 answer:
kvv77 [185]1 year ago
6 0

The following organic reaction C3H7Br +H2O --> C3H7OH +HBr takes place in aqueous state and forms bromine water.

<h3>What is an organic reaction?</h3>

An organic reaction is which the reaction takes place with hydrocarbon compounds or functional groups like hydroxyl or amine or ketones.

In the organic reaction of C3H7Br +H2O --> C3H7OH +HBr the formation of hydroxyl group takes place and because of aqueous medium bromine water is formed.

Therefore, following organic reaction C3H7Br +H2O --> C3H7OH +HBr takes place in aqueous state and forms bromine water.

Learn more about organic reaction , here:

brainly.com/question/580297

#SPJ1

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Whats heptane's formula?
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Heptane means seven carbon atoms and since the general formula is CnH2n+2, heptane will be C7H16
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How many grams of hydrogen gas would be produced from the use of 9.5 moles of aluminum
DIA [1.3K]

Answer:

Identify one disadvantage to each of the following models of electron configuration:

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Explanation:

Identify one disadvantage to each of the following models of electron configuration:

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How many carbon atoms are found in a molecule with the chemical formula c6h12o6 is probably a carbohydrate and monosaccharide be
valentina_108 [34]
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3 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
3 years ago
What is the product of SO2+H2O. ​
Mariulka [41]

Answer:

H2O + SO2 → H2SO3

Explanation:

H2O + SO2 → H2SO3

6 0
3 years ago
Read 2 more answers
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