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:
0.0847M is molarity of sodium hydrogen citrate in the solution
Explanation:
The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.
<em>Moles Na2C6H6O7:</em>
<em>Molar Mass:</em>
2Na: 2*22.99g/mol: 45.98g/mol
6C: 6*12.01g/mol: 72.01g/mol
6H: 6*1.008g/mol: 6.048g/mol
7O: 7*16g/mol: 112g/mol
45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol
Moles of 2g:
2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>
<em />
<em>Liters solution:</em>
100mL * (1L / 1000mL) = <em>0.100L</em>
<em>Molarity:</em>
8.473x10⁻³ moles / 0.100L =
<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
To determine the mass of xenon tetrafluoride, we need to know first the number of fluorine atoms present in <span>oxygen difluoride. We need to convert first the mass into moles then make use of the relation of the elements from the chemical formula. Then, use the avogadro's number to convert it to number of atoms. Then, we do the reverse of the steps above but this time for </span><span>xenon tetrafluoride.
25.0 g OF2 ( 1 mol / 54 g ) ( 2 mol F / 1 mol OF2 ) ( 6.022 x10^23 atoms F / 1 mol F ) ( 1 mol / 6.022x10^23 atoms) ( 1 mol XeF4 / 4 mol F ) (207.3 g / 1 mol XeF4) = 47.99 g XeF4</span>
Answer:2.55 x 10^23 molecules
Explanation:
i had the same question for my class lol, do u go to forest park?
Energy I think that is the answer.
