Answer:
22.9 Liters CO(g) needed
Explanation:
2CO(g) + O₂(g) => 2CO₂(g)
? Liters 32.65g
= 32.65g/32g/mol
= 1.02 moles O₂
Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)
∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)
Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.
∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed
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*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).
Answer : The concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is 
and 
is excess reagent.
First we have to calculate the moles of KSCN.
Moles of KSCN = Moles of 
= Moles of = 
Now we have to calculate the concentration of
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of is,
I believe it is A because the plant is offering protection while the bacteria is converting it in to nitrogen that the plant can use!
First, in order to calculate the specific heat capacity of the metal in help in identifying it, we must find the heat absorbed by the calorimeter using:
Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g
Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.
-279.45 = 50 * c * (11.08 - 45) [minus sign added as energy released]
c = 0.165
The specific heat capacity of the metal is 0.165 cal/gC
Hey there!:
V1 = 3.05 L
V2 = 3.00 L
P1 = 724 mmHg
P2 = to be calculated
T1 = 298 K
T2 = 273 K
Therefore:
P1*V1 / T1 = P2*V2 / T2
P2 = ( P1*V1 / T1 ) * T2 / V2
P2 = 724 * 3.05 * 273 / 298 * 3.00
P2 = 602838.6 / 894
P2 = 674.31 mmHg
1 atm ----------- 760 mmHg
atm ------------- 674.31 mHg
= 674.31 * 1 / 760
= 0.887 atm
Hope this helps!
