Temperature change, colour change, releasing gas, bubbles and change in odor
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
213034 torr is the osmotic pressure.
Explanation:
osmotic pressure is calculated by the formula:
osmotic pressure= iCrT
where i= no. of solute
c= concentration in mol/litre
R= Universal Gas constant
T = temp
It is given that solution is 3% which is 3gms in 100 ml.
let us calculate the concentration in moles/litre
3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl
= 5.372 gm/litre
Putting the values in the formula, Temp in Kelvin 318.5K
osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083
= 284.023 bar or 213018 torr. c= 5.372 moles/L
i=2 for NaCl
The answer is Electron
Hope this helped!!! :)
The balanced chemical equation would be as follows:
<span>K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)
We are given the amount of </span>K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:
65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced
