Answer:
0.26×10²³ molecules
Explanation:
Given data:
Volume of gas = 1.264 L
Temperature = 168°C
Pressure = 946.6 torr
Number of molecules of gas = ?
Solution:
Temperature = 168°C (168+273= 441 K)
Pressure = 946.6 torr (946.6/760 = 1.25 atm)
Now we will determine the number of moles.
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.25 atm ×1.264 L / 0.0821 atm.L/ mol.K ×441 K
n = 1.58 /36.21 /mol
n = 0.044 mol
Now we will calculate the number of molecules by using Avogadro number.
1 mol = 6.022×10²³ molecules
0.044 mol × 6.022×10²³ molecules/ 1mol
0.26×10²³ molecules
Exothermic reactions is where energy is released. Exothermic reactions are reactions that release energy into the environment in the form of heat. Exothermic reactions feel warm or hot or may even be explosive.
Answer:
Concentration of chloride ions = 0.584M
Explanation:
The step by step calculations is shown as attached below.
Answer:
Explanation:
Mass of compound A = 25g
Mass of compound B = 40g
Mass of final mixture = 55g
What happens to the missing mass?
According to the law of conservation of mass, in chemical reaction, matter is transformed from one form to another but cannot be created nor destroyed.
We expect the final mass of the mixture and that of the reacting compounds to be the same but the opposite is the case.
There is a mass loss which typifies most chemical reaction.
The reason for this is that some of the masses must have been lost by the production of gaseous species which are unaccounted for.
The missing mass:
Total mass expected = mass of A + mass of B = 25 + 40 = 65g
Missing mass = expected mass - mass of final mixture = 65 - 55 = 10g
The balanced chemical reaction is:
<span>C4H8O2 + 2H2--> 2C2H6O
</span>A. How many grams of ethyl alcohol are produced by reaction of 2.7 mol of ethyl acetate with H2? 2.7 mol C4H8O2 ( 2 mol C2H6O / 1 mol C4H8O2) (46.07 g / 1 mol) = 248.78 g ethyl alcohol
B. How many grams of ethyl alcohol are produced by reaction of 13.0g of ethyl acetate with H2?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol C2H6O / 1 mol C4H8O2) (<span>46.07 g / 1 mol) = 13.59 g ethyl alcohol</span>
C. How many grams of H2 are needed to react with 13.0g of ethyl acetate?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol H2 / 1 mol C4H8O2) (2.02 g / 1 mol) = 0.5961 g H2
