Answer:
5.12 molecules of Ag₂S could be produced from 10 molecules of silver.
Explanation:
Given data:
Molecules of Ag = 10 molecules
Molecules of Ag₂S = ?
Solution:
First of all we will convert the number of molecules into gram.
10/ 6.022 ×10²³ = 1.7 × 10⁻²³ mol
Mass of Ag:
Number of moles = mass/ molar mass
Mass = Number of moles × molar mass
Mass = 1.7 × 10⁻²³ mol × 107.9 g/mol
Mass = 183.43 × 10⁻²³ g
Chemical equation:
2Ag + H₂S → Ag₂S + H₂
Now we will compare the moles of silver with Ag₂S from balanced chemical equation.
Ag : Ag₂S
2 : 1
1.7 × 10⁻²³ : 1/2 ×1.7 × 10⁻²³ = 0.85 × 10⁻²³ mol
Molecules of Ag₂S:
0.85 × 10⁻²³ × 6.022 ×10²³ = 5.12 molecules
Answer:
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Explanation:
<u>1. Calculate the number of moles of NiCl₂</u>
a) Identify the known variables
- V =0.800 liters
- M = 0.531M
- n = ?
b) Formula
c) Clear n
<u>g) Substitute and compute</u>
<u>2. Convert moles to grams</u>
a) molar mass of NiCl₂
- 129g/mol (shown in the problem)
b) Unit canceling method
Use the factors in order to cancel the units to obtain grams
Note that the mol unit appears on the numerator and denominator, so it cancels leaving just g (grams).
c) Round to 3 signficant figures
Volume = length × width × height
30 = 2 × 3 × h
30 = 6h
5=h
It is 5 cm high
Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
