Answer:
A) 2.76 x 103 kJ
Explanation:
CH3OH (I)--------->CO (g) + 2H2 (g)
Number of moles contained in 87.1g of hydrogen gas= mass of hydrogen gas/ molar mass of hydrogen gas
Molar mass of hydrogen gas= 2gmol-1
Number of moles hydrogen gas = 87.1g/2gmol-1= 43.55 moles of hydrogen
1 mole of methanol yields 2 moles of hydrogen
x moles of methanol yields 43.55 moles of hydrogen
x= 43.55 moles of hydrogen × 1 mole of methanol/2 moles of hydrogen
x= 21.775 moles of methanol
Then;
If 1 mole of methanol absorbs 128KJ of energy
21.775 moles of methanol will absorb 21.775 × 128/1 = 2.7×10^3 KJ of heat
Answer:
There are 
Explanation:
In this problem, we need to find the number of molecules in 
mol of 
.
The molar mass of is 
No of moles = mass/molar mass
We can find mass from above formula.
Also,
No of moles = no of molecules/Avogadro number
Hence, there are
Sally is struggling with Prospective memory.
<u>Explanation:</u>
The type of memory in which an individual has problem to remember a planned set of actions. He or she can remember or recall it only after some period of time may be in future and not at that time where those actions needs to be performed. These type of memory problem can range from a simple situation like the tasks that are to be done in common, simple and daily basis to serious issue.
Some of the instances of this kind of prospective memory includes, closing a tap after usage, cut a call over phone when the conversation is done,etc. These instances includes only simple issues. Some of dangerous issues includes forgetting to take regular medicines, forgetting to take some protective and safety measures while driving,etc.
Answer:
k is 3,18*10⁻² s⁻¹ at 75°C
Explanation:
following Arrhenius equation:
k= k₀*e^(-Ea/RT)
where k= rate constant , k₀= frequency factor , Ea= activation energy , R= universal gas constant T=absolute temperature
then for T₁=25°C =298 K
k₁= k₀*e^(-Ea/RT₁)
and for T₁=75°C = 348 K
k₂= k₀*e^(-Ea/RT₂)
dividing both equations
k₂/k₁= e^(-Ea/RT₂+Ea/RT₁ )
k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )]
replacing values
k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )] = 4,7*10⁻³ s⁻¹ *e^[-33.6*1000 J/mol /8.314 J/molK*(1/ 348 K -1/298 K )] = 3,18*10⁻² s⁻¹
thus k is 3,18*10⁻² s⁻¹ at 75°C
Answer:
1.7700×10^-22 g
Explanation:
A face-centered-cubic cell contains 4 atoms, so the volume per atom is ...
(389.80×10^-12 m)³/(4 atoms) ≈ 1.4725×10^-29 m³
Then the mass of 1 atom is ...
(12020 kg/m³)(1.4725×10^-29 m³) ≈ 1.76995×10^-25 kg
≈ 1.7700×10^-22 g
_____
<em>Additional comment</em>
That is the approximate mass of a Palladium atom.
