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Travka [436]
3 years ago
15

Item 4 part a if a proton and an electron are released when they are 4.50×10−10 m apart (typical atomic distances), find the ini

tial acceleration of each of them.
Chemistry
1 answer:
Wewaii [24]3 years ago
6 0

Acceleration is defined as the rate of change of velocity with respect to time.

Formulas of force are given by:

F=ma   (1)

where,

F = force

m = mass

a = acceleration

F=\frac{kq_{1}q_{2}}{r^{2}}   (2)

where,

F = force

k = coulomb constant (9\times 10^{9}Nm^{2}C^{-2})

r = distance between the charged particles

q_{1} and q_{2} are the signed magnitude of charges

Use the formula (2) for calculating the value of force, we get:

Substitute the value of q_{1} and q_{2}, k and r to find the value of force.

F= \frac{9\times 10^{9} Nm^{2}C^{-2}\times (1.6\times 10^{-19} C)^{2}}{(4.50\times 10^{-10})^{2}}

= 1.1378\times 10^{-9} N

Now, put the above value force in formula (1) to identify the initial acceleration.

1.1378\times 10^{-9} N=ma   (1)   (mass of electron  =9.1 \times 10^{-31} kg)

acceleration of electron = \frac{1.1378\times 10^{-9} N}{9.1\times 10^{-31}kg}

= 1.25\times 10^{21}N/kg

And,

acceleration of proton =\frac{1.1378\times 10^{-9} N}{mass of proton}

mass of proton =1.67 \times  10^{-27}

Thus,

acceleration of proton = \frac{1.1378\times 10^{-9} N}{1.67 \times  10^{-27} kg}

=6.81 \times  10^{17} N/kg

Now, initial acceleration of electron and proton is  1.25\times 10^{21}m/s^{2} and 6.81 \times  10^{17} m/s^{2} as 1 N = kgm/s^{2}.

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HURRY PLEASE HELP
Dahasolnce [82]

Answer:

A

Explanation:

its A

facts o.o

7 0
3 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
3 years ago
Butanal, an aldehyde, can be made from 1-bromobutane, but it requires two reactions in sequence. What are the two reagents that
hammer [34]

Answer:

See explanation and image attached

Explanation:

My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.

1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.

These sequence of reactions are shown in the image attached.

4 0
3 years ago
BRAINLIEST TO FIRST CORRECT ANSWER, THANKS!
Tamiku [17]

Answer:

increased

Explanation:

Consuming a compound increases the concentration. When you increase the concentration, the rate constant for that reaction also increases.

5 0
3 years ago
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