Given :
Mass of gold ( Au ) is 212 gram.
To Find :
Atoms of gold present in 212 gram of gold.
Solution :
Molecular mass of gold is 197 gram.
So, number of moles of gold in 212 gram is :
Now, we know 1 mole of any element contains atoms.
So, number of atoms present in 1.076 moles are :
Hence, this is the required solution.
Answer:
30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.
Explanation:
The reaction that takes place is:
Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:
0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>
Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:
Finally we convert Mg(OH)₂ moles to grams:
Answer:
The answer to your question is: T2 = 235.44 °K
Explanation:
Data
V1 = 3.15 L V2 = 2.78 L
P1 = 2.40 atm P2 = 1.97 atm
T1 = 325°K T2 = ?
Formula
Process
T2 = (P2V2T1) / (P1V1)
T2 = (1.97x 2.78x 325) / (2.40 x 3.15)
T2 = 1779.895 / 7.56
T2 = 235.44 °K