Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl
According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]
and we have [Br2] = 0.00366 m [Cl2]= 0.000672 m [BrCl] = 0.00415 m
by substitution:
= [0.00415]^2 / ( [0.00366] * [0.000672])
∴ Q = 7
4.4moles of oxygen atoms
Explanation:
Given parameters:
Mass of MgSO₄ = 132.2g
Unknown:
Number of moles of oxygen atoms = ?
Solution:
The number of moles is the quantity of substance that contains the avogadro's number of particles.
To solve for this;
Number of moles = 
Molar mass of MgSO₄ = 24 + 32 + 4(16) = 120g/mole
Number of moles = 
= 1.1 moles
In
1 moles of MgSO₄ we have 4 moles of oxygen atoms
1.1 moles of MgSO₄ contains 4 x 1.1 moles = 4.4moles of oxygen atoms
learn more:
number of moles brainly.com/question/1841136
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<span>Let's assume
that the F</span>₂ gas has ideal gas behavior.
<span>
Then we can use ideal gas formula,
PV = nRT
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.</span>
Moles = mass / molar mass
Molar mass of F₂ = 38 g/mol
Mass of F₂ = 76 g
Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol
<span>
P = ?
V = 1.5 L = 1.5 x 10</span>⁻³ m³
n = 2 mol
R = 8.314 J mol⁻¹ K⁻<span>¹
T = -37 °C = 236 K
By substitution,
</span>
P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K
p = 2616138.67 Pa
p = 25.8 atm = 26 atm
Hence, the pressure of the gas is 26 atm.
Answer is "a".
<span>
</span>
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
ANSWER: B. 20 grams since no matter was added or removed
Hope it helps!
