Aye you have the same class as me bruh I need help on some chemistry qustions
Answer:
It is made up of two elements.
Explanation:
To answer the question given above,
We shall determine the number of elements present in CaCl₂. This can be obtained as follow:
CaCl₂ contains calcium (Ca) and chlorine gas (Cl₂).
This implies that CaCl₂ contains two different elements.
Now, considering the options given in the question above, CaCl₂ is made up of two elements.
MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
MgBr2(s) --> Mg+(aq) + 2 Br⁻(aq)
[Br⁻] = 0.065 mol MgBr2/1L × 2 mol Br⁻ / 1 mol MgBr2 = 0.13 M
The answer to this question is [Br⁻] = 0.13 M
Answer:
<h3>The answer is 7.42 </h3>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ { H_3O}^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7B%20H_3O%7D%5E%7B%2B%7D%5D)
From the question we have

We have the final answer as
<h3>7.42 </h3>
Hope this helps you
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.