Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
When a chemical reaction takes place it changes the composition of the reactants. The ways to tell if a reaction is occurring are,
1-Release of heat
2-Production of a gas
3-Formation of a precipitate
4-Change in color
The average speed : 3.07 m/s
<h3>Further explanation</h3>
Given
d1 = 2 km
t1 = 9 min
d2 = 1.5 km
t2 = 10 min
Required
The average speed
Solution
Average speed = total distance : total time
total distance :
= d1+d2
= 2+1.5
=3.5 km
total time :
= t1+t2
= 9 min + 10 min
= 19 min
Average speed :
= 3.5 km : 19 min
= 0.184 km/min
= 3.07 m/s
Answer:
Molar concentration of S₂ is 1.77×10⁻⁶M
Explanation:
For the reaction:
2H₂S(g) ⇄ 2H₂(g) + S₂(g)
The equilibirum constant, K, is defined as:
![K = \frac{[S_2][H_2]^2}{[H_2S]^2}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BS_2%5D%5BH_2%5D%5E2%7D%7B%5BH_2S%5D%5E2%7D)
<em>(1)</em>
Concentrations in equilibirum are:
[H₂S] : 0,163/0.500L - X
[H₂] : 0,0500/0.500L + X
[S₂] : X
Replacing the concentrations and the equilibrium value in (1):
![K = \frac{[X][0.1+X]^2}{[0326-X]^2}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BX%5D%5B0.1%2BX%5D%5E2%7D%7B%5B0326-X%5D%5E2%7D)
1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)
1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X
0 = X³ + 0.2X² + 0.01X - 1.77x10⁻⁸
Solving for X:
X = 1.77×10⁻⁶
As [S₂] = X, <em>molar concentration of S₂ is 1.77×10⁻⁶M</em>
I hope it helps!
<h2><u><em>D:Water enters an aquifer throught infitation </em></u></h2>
