Answer:
130 g of sucrose
Explanation:
Boiling point elevation formula → ΔT = Kb . m
ΔT = Boiling T° solution - Boiling T° pure solvent → 0.39°C
0.39°C = 0.513°C/m . M
m = 0.760 mol/kg → molality = moles of solute / 1kg of solvent
Let's determine the moles of solute → molality . kg
0.760 mol/kg. 0.5 kg = 0.380 moles
If we convert the moles to mass, we'll get the answer
0.380 mol . 342.30 g/mol = 130g
Answer:
1s22s22p6: Neon (Ne)
1s22s22p63s23p3: Phosphorous (P)
1s22s22p63s23p64s1: Potassium (K)
1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)
1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4
Explanation:
you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number
First we must write a balanced chemical equation for this reaction

The mole ratio for the reaction between
and
is 1:2. This means 1 moles of
will neutralize 2 moles
. Now we find the moles of each reactant based on the mass and molar mass.



The
was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.
Answer:
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Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

And the resulting concentration of KOH and OH ions as this is a strong base:
![[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5Cfrac%7B0.00576mol%7D%7B0.012L%2B0.032L%7D%3D0.131M)
And the resulting pH is:

Regards!