Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
<span>Answer:
H-C-N H-N-C C-H-N
Notice that C-H-N is the same as N-H-C just written backwards. ( i.e. they have the same connectivtiy.) You can exclude the last one with H in the middle since H has two bonds and 4 electrons around it. At this point you couldn't differentiate between the first two, so I would give you the connectivity in such a problem, which in this case is H-C-N.</span>
Answer:
Mg2 + O2 → 2MgO
Explanation:
Hope this helps!! I got it right.
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
