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3 years ago

Std 10th

Chemistry

1 answer:

kondaur [170]3 years ago

5 0

Answer:

zinc and lead or copper and tin

Explanation:

these elements react both as an acid as well as a base

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At the end of the reaction you washed the chemicals into a beaker using water. Why was it OK to use water to do this if Grignard

lesya692 [45]

Answer: water could be used to wash it since the reaction has ended.

Explanation:

There will be no reaction of water with the Grignard reagent since the reaction has ended, as it is well known that water is a universal solvent for washing of glasswares after experiments but if it is during the reaction it will be more advisable to rinse with alcohol to enhance more accuracy during the experiment

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3 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

6 0

3 years ago

What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?

natita [175]

This question provides us –

Weight of $\bf KCl$ is = 47 g
Volume, V = 375 mL

__________________________________________

Molar Mass of $\bf KCl$ –

$\qquad$ $\twoheadrightarrow\bf 39.0983 \times 35.453$

$\qquad$ $\twoheadrightarrow\bf 74.5513$

<u>Using formula</u> –

$\qquad$ $\purple{\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ W\times 1000}{MV}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ 47 \times 1000}{74.5513\times 375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{47000}{27956.7375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \cancel{\dfrac{47000}{27956.7375}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = 1.68117M$

$\qquad$ $\pink{\twoheadrightarrow\bf Molarity _{(Solution)} = 1.7M}$

Henceforth, Molarity of the solution is = 1.7M

___________________________________________

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2 years ago

When sugar dissolves in water are covalent bonds broken?

Amanda [17]

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3 years ago

Lots of points for help!!!!

Svetllana [295]

Answer:

a+b=C

Explanation:

8 0

3 years ago