Answer: 0,4278g of F and 0,4191g of Fe
Explanation: it's possible to calculate the mass of each element by multiplying the percentage (decimal) of the element by the mass of the compound.
For Fluorine (F)
0,847g * 0,5051 = 0,4278g of F
For iron (Fe)
0,847 * 0,4949 = 0,4191g of Fe
This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.
At the end, the sum of the elements masses should be the total mass of the compound.
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)
so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
Answer:
Four moles of the cation
Explanation:
2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)
Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.
The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.
This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.
The concentration of the solution is 4.25 M
Explanation
molarity=moles/volume in liters
moles = mass/molar mass
molar mass of HF = 19 + 1 = 20 g/mol
moles is therefore = 17.0 g/ 20 g/mol = 0.85 moles
volume in liters = 2 x10^2ml/1000 = 0.2 liters
therefore molarity = 0.85/0.2 = 4.25 M
