I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is :
(where Q is heat, m is mass, c is specific heat and
is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature
Answer:
Explanation:
To convert from moles to grams, the molar mass must be used.
1. Find Molar Mass
The compound is iron (III) chloride: FeCl₃
First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).
There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.
- FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol
This number tells us the grams of FeCl₃ in 1 mole.
2. Calculate Moles
Use the number as a ratio.
Multiply by the given number of grams, 345.0.
Flip the fraction so the grams of FeCl₃ will cancel.
Divide.
3. Round
The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.
For the answer we calculated, that is the thousandth place.
The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.
There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.
Answer:
what's the question? is it A
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
