MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
Answer:-
molecules.
Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.
It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.
In second step, the moles are converted to molecules on multiplying by Avogadro number.
Molar mass of
= 12+4(79.9) = 331.6 g per mol
let's make the set up using dimensional analysis:

=
molecules
So, there will be
molecules in 250 grams of
.
Smaller atoms ; free neutrons and energy
When we wish to convert a gas to liquid we have to either
a) decrease temperature
b) increase pressure
In case of fire extinguisher the CO2 is found to be in liquid state, this is as the CO2 is pressurized at high pressure which keeps CO2 in liquid state
the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas
Pressure = 5 - 73 atm
Temperature = -57 to 31 degree Celsius
Answer:
The enthalpy change for the given reaction is 424 kJ.
Explanation:

We have :
Enthalpy changes of formation of following s:



(standard state)
![\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5B%5CDelta%20H_f%28product%29%5D-%5Csum%20%5B%5CDelta%20H_f%28reactant%29%5D)
The equation for the enthalpy change of the given reaction is:
=

=


The enthalpy change for the given reaction is 424 kJ.