Answer:
You need to dissolve 144 gr of glucose in 0.8 liters of water to have a 1 M solution.
Explanation:
Molarity is the amount of moles per liter of solution --> moles/liters. We are trying to get a 1 molar solution and we also know the liters (0,8L) so
![1\frac{Moles-of-glucose}{liters-of-water } * 0.8 liters of water](https://tex.z-dn.net/?f=1%5Cfrac%7BMoles-of-glucose%7D%7Bliters-of-water%20%7D%20%2A%200.8%20liters%20of%20water)
after canceling liters of water you get 0,8 moles of glucose.
![180 \frac{grams-of-glucose}{mol-of-glucose}*0.8 mol- of- glucose=144 grams-of-glucose](https://tex.z-dn.net/?f=180%20%5Cfrac%7Bgrams-of-glucose%7D%7Bmol-of-glucose%7D%2A0.8%20mol-%20of-%20glucose%3D144%20grams-of-glucose)
Answer:
Kc = 4.76
Explanation:
To find the concentrations of CO and H₂ at equilibrium, you have to set up an ICE (Initial, Change, Equilibrium) table.
CO (g) + 2H₂(g) ⇌ CH₃OH (g)
I 0.32 M 0.53 M 0
C -x -2x +x
E 0.32-x 0.53-2x 0.16 M
Since you know the concentration of CH₃OH at equilibrium, it would be equal to x since 0 + x = 0.16. So,
[CH₃OH] = 0.16 M
[CO] = 0.32 - 0. 16 = 0.16 M
[H₂] = 0.53 - 2(0.16) = 0.21 M
Now that you have all the concentrations at equilibrium, you can calculate the equilibrium constant.
Kc = products ÷ reactants
= [CH₃OH] ÷ [CO][H₂]²
= 0.16 ÷ (0.16)(0.21)
Kc = 4.76
The equilibrium constant at this temperature is 4.76.
Hope this helps.
I believe the answer is B
I think of my backyard. There's tons and tons of flowers back there that attracts many bees during the spring and summer. The bee and the flower have a mutualistic relationship, and here's how: the bee goes from flower to flower, getting pollen, which benefits them. Some of this pollen gets on their bodies, and when they go to another flower, said pollen can rub off on other flowers, giving those flowers pollen. I hope this makes sense!
Answer:
![X_{H_2}=\0.3584](https://tex.z-dn.net/?f=X_%7BH_2%7D%3D%5C0.3584)
![X_{CO_2}=\0.6416](https://tex.z-dn.net/?f=X_%7BCO_2%7D%3D%5C0.6416)
Explanation:
According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.
Partial pressure of carbon dioxide = 401 mmHg
Partial pressure of hydrogen gas = 224 mmHg
Total pressure,P = sum of the partial pressure of the gases = 401 + 224 mmHg = 625 mmHg
Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.
So,
![P_{CO_2}=X_{CO_2}\times P](https://tex.z-dn.net/?f=P_%7BCO_2%7D%3DX_%7BCO_2%7D%5Ctimes%20P)
![X_{CO_2}=\frac{P_{CO_2}}{P}](https://tex.z-dn.net/?f=X_%7BCO_2%7D%3D%5Cfrac%7BP_%7BCO_2%7D%7D%7BP%7D)
![X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}](https://tex.z-dn.net/?f=X_%7BCO_2%7D%3D%5Cfrac%7B401%5C%20mmHg%7D%7B625%5C%20mmHg%7D)
![X_{CO_2}=\0.6416](https://tex.z-dn.net/?f=X_%7BCO_2%7D%3D%5C0.6416)
Similarly,
![X_{H_2}=\frac{P_{H_2}}{P}](https://tex.z-dn.net/?f=X_%7BH_2%7D%3D%5Cfrac%7BP_%7BH_2%7D%7D%7BP%7D)
![X_{H_2}=\frac{224\ mmHg}{625\ mmHg}](https://tex.z-dn.net/?f=X_%7BH_2%7D%3D%5Cfrac%7B224%5C%20mmHg%7D%7B625%5C%20mmHg%7D)
![X_{H_2}=\0.3584](https://tex.z-dn.net/?f=X_%7BH_2%7D%3D%5C0.3584)