potassium chromate - k2cro4. ,,silver nitrate-agno3. ,, potassium nitrate-kno3 ,,Silver chromate- ag2cro4
Answer:
28%
Explanation:
Basically, all o did was write the equations, balance it and solve for them. Also, at the place I stared, I used simultaneous equation to solve it. Multiplying by 8 and also 3.
It's a pretty straightforward question.
At the final step that's missing, I Did
(y)C3H8 = 2.8 / ( 2.8 + 7.1)
(y)C3H8 = 0.28
Answer:
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Explanation:
ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.
In our case we have:
mass of MgBr₂ = 12.41 g
volume of water (which is equal to the final solution volume) = 2.55 L
Now we devise the following reasoning:
if 12.41 g of MgBr₂ are dissolved in 2.55 L of water
then X g of MgBr₂ are dissolved in 1 L of water
X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂
if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻
then in 4.867 g of MgBr₂ we have Y g of Br⁻
Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)
4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
![Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B802%5C%2C%5Cfrac%7BkJ%7D%7Bmol%7D%20%7D%7B16%5C%2C%5Cfrac%7Bg%7D%7Bmol%7D%20%7D)
![Q = 50.125\,\frac{kJ}{g}](https://tex.z-dn.net/?f=Q%20%3D%2050.125%5C%2C%5Cfrac%7BkJ%7D%7Bg%7D)
1 gramo de metano aporta 50.125 kilojoules.
Octano
![Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B5500%5C%2C%5Cfrac%7BkJ%7D%7Bmol%7D%20%7D%7B114%5C%2C%5Cfrac%7Bg%7D%7Bmol%7D%20%7D)
![Q = 48.246\,\frac{kJ}{mol}](https://tex.z-dn.net/?f=Q%20%3D%2048.246%5C%2C%5Cfrac%7BkJ%7D%7Bmol%7D)
1 gramo de metano aporta 48.246 kilojoules.
Because it contains vinegar because it does not form layers when mixed with other liquids. Sugar or citric acid because they don't leave sediment.