Answer:
they don't depend on the temperature
Answer:
49.5J/°C
Explanation:
The hot water lost some energy that is gained for cold water and the calorimeter.
The equation is:
Q(Hot water) = Q(Cold water) + Q(Calorimeter)
<em>Where:</em>
Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J
Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J
That means the heat gained by the calorimeter is
Q(Calorimeter) = 4794J - 3834J = 960J
The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:
59.4°C-40°C = 19.4°C
And calorimeter constant is:
960J/19.4°C =
<h3>49.5J/°C</h3>
<em />
Answer:
Hydrogen ion concentration is 2.773E-4Mole per DM3
Hydroxide ion concentration is 4.093E-11
It is an acid
Explanation:
PH(H3O)=-LogH3O
PH=3.557
H3O=antilog of -3.557
antilog of -3.557=antilog (0.443 -4)
antilog0.443 ×antilog -4=2.773E-4mole per DM3.
(H3O)(OH)=10^-14
H30=2.773×10^-14
OH= 1.00E-14/2.773E-4
=4.093E-11
it is an acid since the concentration of hydrogen ion is more than the hydroxide ion
Answer:
They're too small
Explanation:
Atoms are smaller than the wavelength of visible light
Answer:
The concentration of the standard NaOH solution is 0.094 moles/L.
Explanation:
In the titration, the equivalence point is defined as the point where the moles of NaOH (the titrant) and KHP (the analyte) are equal:
moles of NaOH = moles of KHP
![[NaOH]xV_{NaOH} = moles of KHP](https://tex.z-dn.net/?f=%5BNaOH%5DxV_%7BNaOH%7D%20%3D%20moles%20of%20KHP)
![[NaOH] = \frac{moles of KHP}{V_{NaOH}}](https://tex.z-dn.net/?f=%5BNaOH%5D%20%3D%20%5Cfrac%7Bmoles%20of%20KHP%7D%7BV_%7BNaOH%7D%7D)
The 
is 40.82mL = 0.04082L and the moles of KHP are
Replacing at the first equation:
![[NaOH] = \frac{3.827x10^{-3}moles}{0.04082L} = 0.094 moles/L](https://tex.z-dn.net/?f=%5BNaOH%5D%20%3D%20%5Cfrac%7B3.827x10%5E%7B-3%7Dmoles%7D%7B0.04082L%7D%20%3D%200.094%20moles%2FL)
