Answer:
The answer is in the explanation.
Explanation:
The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.
The reaction that explain this use is:
KHP + NaOH → KNaP + H2O
<em>where 1 mole of KHP reacts per mole of NaOH</em>
That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added
With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.
The moles of KHP are obtained from the volume and the concentration as follows:
Volume(L)*Concentration (Molarity,M) = moles of KHP
If the concentration is more or less than 0.100M, the moles will be higher or lower. For that reason, we need to know the concentration of KHP but is not necessary to be 0.100M.
Answer:
76,6 kg
Explanation:
A kg it's equal to 1x10^3 grams
A Gigagrams it's equal to 1x10^9 grams
Knowing this, a kg it's equal to 1x10^6 gigagrams
![7,66*10^{-5}[gigagram]*\frac{1*10^6 [kg]}{1 [gigagram]}= 76.6 [kg]](https://tex.z-dn.net/?f=7%2C66%2A10%5E%7B-5%7D%5Bgigagram%5D%2A%5Cfrac%7B1%2A10%5E6%20%5Bkg%5D%7D%7B1%20%5Bgigagram%5D%7D%3D%2076.6%20%5Bkg%5D)
Answer:
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Explanation:
ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm.................... candyunicorns1999 has left the chat
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
