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Svetllana [295]
3 years ago
10

A professional technician maintains and operates specialized scientific equipment true or false

Chemistry
1 answer:
tino4ka555 [31]3 years ago
4 0
True for both academic and industrial environments.<span />
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The Everglades and the Louisiana wetan are the same
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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 0.70

Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL

Hence, the volume of concentrated solution required is 9.95 mL

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2 years ago
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Explain why 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane. hint: examine the
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Answer:-

The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.

The I - attacks from backside to give the transition state for both.

If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.

This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.

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Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2
Zinaida [17]

Explanation:

Expression for the kinetic energy is as follows.

         K.E = \frac{1}{2}mv^{2}

Now, total kinetic energy will be as follows.

    K.E = 2 \times \frac{1}{2}mv^{2} = m \times v^{2}

Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

   Energy of photon = \frac{hc}{\lambda}

                                = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}

                                 = 1.63 \times 10^{-18} J

    m \times v^{2} = 1.63 \times 10^{-18}

          v = \sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}

             = 3.12 \times 10^{4} m/s

Thus, we can conclude that atoms were moving at a speed of 3.12 \times 10^{4} m/s before the collision.

8 0
3 years ago
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