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3 years ago

What is the velocity of a electron whose wavelength is 100 pm?

Chemistry

1 answer:

Anni [7]3 years ago

5 0

Answer:

v = 7.3 × 10⁶ m/s

Explanation:

Given data:

Velocity of electron = ?

Wavelength = 100 pm

Solution:

Formula:

λ = h/mv

λ = wavelength

h = planck's constant

m = mass

v = velocity

Now we will put the values in formula.

100 ×10⁻¹² m = 6.63 × 10⁻³⁴ j.s / 9.109 × 10⁻³¹ kg × v

v = 6.63 × 10⁻³⁴ kg.m²/s / 9.109 × 10⁻³¹ kg ×100 ×10⁻¹² m

v = 6.63 × 10⁻³⁴ m/s /910.9 × 10⁻⁴³

v = 0.0073 × 10⁹ m/s

v = 7.3 × 10⁶ m/s

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How many moles of gas would occupy a 25.0 liter when the temp is 22.0 degrees celsius and the pressure is 646 torr

Luda [366]

Answer:

0.877 mol

Step-by-step explanation:

We can use the Ideal Gas Law to solve this problem.

pV = nRT Divide both sides by RT

n = (pV)/(RT)

Data:

p = 646 torr

V = 25.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 22.0 °C

Calculations:

(a) Convert the pressure to atmospheres

p = 646 torr × (1 atm/760 torr) = 0.8500 atm

(b) Convert the temperature to kelvins

T = (22.0 + 273.15) K = 295.15 K

(c) Calculate the number of moles

n = (0.8500 × 25.0)/(0.082 06 × 295.15)

= 0.877 mol

3 0

3 years ago

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago

If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut

Aneli [31]

Hey There!

At neutralisation moles of H⁺ from HCl = moles of OH⁻ from Ca(OH)2 so :

0.204 * 42.8 / 1000 => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles ( since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass = 0.0043656 * 74.1

mass = 0.32 g of Ca(OH)2

Hope that helps!

6 0

4 years ago

The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use EscE 0.241V.

tino4ka555 [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $K_b = 1.89 *10^{-6}$

Explanation:

From the question we are told that

The voltage of the cell is $V = 0.731 \ V$

Generally $K_b$ is mathematically represented as

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

Where $K_w$ is the equilibrium constant for this auto-ionization of water with a value $K_w = 1.0 *10^{-14}$

Generally the $E_{cell}$ is mathematically represented as

$E_{cell} = V - E_{SCE}$

=> $E_{cell} = 0.731 - 0.241$

=> $E_{cell} = 0.49 V$

This $E_{cell}$ is mathematically represented as

$E_{cell} = \frac{0.0592}{n} * log K_a$

Where n is the number of moles which in this question is n = 1

$0.490 = \frac{0.0592}{1} * log K_a$

=> $K_a = 5.30*10^{-9}$

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

=> $K_b = \frac{1.0 *10^{-14}}{ 5.30*10^{-9}}$

=> $K_b = 1.89 *10^{-6}$

5 0

3 years ago

M. element cenabrecht 0.5 mele of an

drek231 [11]

Answer:

Number of moles (n)=

molecular weight

weight

Weight=n×Molecular weight

=0.5×14

Mass=7g

5 0

3 years ago