Answer:
Explanation:
So, the formula for the compound should be:

Now we assume that we have 1 mol of substance, so we can make calculations to know the molar mass of element X, as follows:

So we have that 6 moles weight 212.7g, and we can make a rule of three to know the weight of compound X:

As we used 1 mol, we know that the molar mass is 32.06g/mol
So the element has a molar mass of 32.06 g/mol and an oxidation state of +6, with this information, we can assure that the element X is sulfur, so the compound is 
Answer: Too much base was added
i guessed
Explanation:
Answer:
2 electrons
Explanation:
There are five 3d orbitals, each of which can hold up to 2 electrons, for 10 total electrons. An orbital is described by the principle quantum number, n, the angular momentum quantum number, l, and the magnetic quantum number, ml.
The equation is: C+O2=>CO2
Since we got 10 molecules of CO2 new balanced equation would be 10C+10O2=>10CO2
from this equation we can see that we have 10 molecules of oxygen, however ,we need to find atoms. There are 2 atoms in the oxygen molecule so we need to multiply 10 by 2 which gives us 20 atoms.
The answer: there are 20 atoms of oxygen
Answer:
11.39
Explanation:
Given that:


Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:

Thus,


Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)


Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)

x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>