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denpristay [2]
3 years ago
15

How to find how many atoms are in grams.

Chemistry
1 answer:
professor190 [17]3 years ago
5 0

Explanation:

To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23.

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A certain box contains 200 particles of an ideal gas. how many times more likely is it to find the particles evenly split betwee
Strike441 [17]

 Each particle has two possibilities: either it is on the left hand side or it is on the right hand side.  

The probability of having them equally split is  

P(100,100) = 200C100 / 2^200  

where 200C100 is the binomial coefficient.  

For the uneven distribution we have  

P= 2 * 200C160 / 2^200  

where the factor 2 comes in because we could have 160/40 or 40/160 as a division.  

So the ratio of probabilities is  

200C100 /(2 * 200C160 )  

= 2.2 * 10^16

Hope this helps!

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3 years ago
Name a Solid, Liquid, and a Gas. Solid: Liquid: Gas:
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3 years ago
Read 2 more answers
When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be
Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH  = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL  +  50.mL  = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH_{neutralization}

ΔH_{neutralization} = -q_soln / mole of water produced

so we substitute

ΔH_{neutralization} = -( 4076.68 J ) / 0.0500 mol  

ΔH_{neutralization} = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0
3 years ago
Describe three different scenarios where a chemical process or reaction shows an increase in entropy.
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