Answer:
0.56 g
Explanation:
<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>
Step 1: Given data
Moles of nitrogen gas (n): 0.020 mol
Step 2: Calculate the molar mass (M) of nitrogen gas
Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:
M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol
Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas
We will use the following expression.
m = n × M
m = 0.020 mol × 28.02 g/mol
m = 0.56 g
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In a crystal, the molecules are closer together as they are in any solid. they have less room to move, and might even be combined together rather than individual
Weathering of the rock and sedimentation are decomposition processes. Through time, the minerals in the rocks soften due to pressure and heat. So, they crumble down and reduce in terms of size. Once they do, they become sand or part of the soil. So, the answer is A.
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
