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Alika [10]
2 years ago
15

What is the de Broglie wavelength (in nm) of a proton of traveling at 3.18% of the speed of light?

Chemistry
1 answer:
gulaghasi [49]2 years ago
7 0
Answer is: <span>de Broglie wavelength of a proton is </span>3,4·10⁻⁵ nm.
v(proton) = 0,038 · 3·10⁸ m/s.
v(proton) = 1,14·10⁷ m/s; speed of proton.
m(proton) = 1,67·10⁻²⁷ kg.
h = 6,62607004·10⁻³⁴ m²·kg/s; Planck constant.
λ(proton) = h / m(proton) · v(proton).
λ(proton) = 6,62607004·10⁻³⁴ m²·kg/s ÷ (1,67·10⁻²⁷ kg · 1,14·10⁷ m/s).
λ(proton) = 3,48·10⁻¹⁴ m · 10⁹ nm/m = 3,4·10⁻⁵ nm.
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If an atom of nitrogen bonded with an atom of aluminum , what kind of bond would form
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2 years ago
Shifts in the rock layer locations cannot account for gaps in the rock record<br><br> true or false
vivado [14]

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Explanation:

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4 0
3 years ago
Write a full chemical equation for the reactions between magnesium and sulfuric acid
melamori03 [73]

Answer:

magnesium + hydrochloric acid → hydrogen gas + magnesium chloride

explanation:

the nitrogen in HNO3 is in the +5 oxidation state and is easily reduced. The reduction would result in the oxidation of the hydrogen gas, forming the water once again.The sulfur in H2SO4 is also in its highest oxidation state, +6.

<em>Hope</em><em> this</em><em> helps</em><em> </em><em>:</em><em>)</em>

6 0
3 years ago
How many O2 molecules react with 47 CH4 molecules according to the preceding equation CH4 + O2 → CO2 + H2O
natka813 [3]
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8 0
3 years ago
Consider the following reaction 2 N2O(g) =&gt; 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, cal
den301095 [7]

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

[N2O} = 0.3325 M

After 2.0 minutes the concentration of N2O is 0.3325 M

3 0
3 years ago
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