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Mars2501 [29]
2 years ago
6

How many moles are in 3.01 x 10^23 atoms of zinc?

Chemistry
1 answer:
Lerok [7]2 years ago
6 0

Answer:

<h2>0.5 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{3.01 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{3.01}{6.02}  \\  = 0.5

We have the final answer as

<h3>0.5 moles</h3>

Hope this helps you

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What is the wavelength of a wave having a frequency of 3.76x10^14
choli [55]

Answer:

3.76 x 1014 s-1? λ = c/ν = 3.00 x 108 m/s = 7.98 x 10-7 m 3.76 x 1014 s-1.

Explanation:

hope i helped btw i am breanna  plzz mark me brainliest

4 0
3 years ago
The specific amount of energy an electron has, or its what, is related to its movement
hram777 [196]
I Think.

Neutron(s).

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5 0
3 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
If nitrogen and hydrogen combine in a combustion reaction, what would the product of the reaction be?
kow [346]

Co2

Explanation:

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

4 0
3 years ago
Read 2 more answers
Na +H₂O- NaOH +H₂ balance
Whitepunk [10]

Hey there!

Na + H₂O → NaOH + H₂

First, balance O.

One on the left, one on the right. Already balanced.

Next, balance H.

Two on the left, three on the right. Let's add a coefficient of 2 in front of NaOH and a coefficient of 2 in front of H₂O, so we have 4 on each side.

Na + 2H₂O → 2NaOH + H₂

Lastly, balance Na.

One on the left, two on the right. Add a coefficient of 2 in front of Na.

2Na + 2H₂O → 2NaOH + H₂  

This is our final balanced equation.

Hope this helps!

7 0
3 years ago
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