Answer:
3.8 M
Explanation:
Volume of acid used VA= 57.0 - 37.5 = 19.5 ml
Volume of base used VB= 67.8 - 45.0 = 22.8 ml
Equation of the reaction
2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)
Number of moles of acid NA= 2
Number of moles of base NB= 1
Concentration of acid CA= ???
Concentration of base CB= 1.63 M
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA= 1.63 × 22.8 × 2/ 19.5 × 1
CA= 3.8 M
HENCE THE MOLARITY OF THE ACID IS 3.8 M.
Answer:
25.11 g.
Explanation:
- It is clear from the balanced equation:
<em>Ag₂O + 2HCl → 2AgCl + H₂O.</em>
<em></em>
that 1.0 mole of Ag₂O reacts with 2.0 moles of HCl to produce 2.0 mole of AgCl and 1.0 moles of H₂O.
- 7.8 g of HCl reacts with excess Ag₂O. To calculate the no. of grams of Ag₂O that reacted, we should calculate the no. of moles of HCl:
<em>no. of moles of HCl = mass/atomic mass</em> = (7.9 g)/(36.46 g/mol) = <em>0.2167 mol.</em>
- From the balanced equation; every 1.0 mol of Ag₂O reacts with 2 moles of HCl.
∴ 0.2167 mol of HCl will react with (0.2617 mol / 2 = 0.1083 mol) of Ag₂O.
<em>∴ The mass of reacted Ag₂O = no. of moles x molar mas</em>s = (0.1083 mol)(231.735 g/mol) = <em>25.11 g.</em>
Answer:
24.309
Explanation:
Let A, B, and C represent the three isotopes.
For isotope A:
Mass number = 23.99
Abundance = 78.99%
For isotope B:
Mass number = 24.99
Abundance = 10.00%
For isotope C:
Mass number = 25.98
Abundance = 11.01%
Atomic mass of Mg =?
Atomic mass = [(Mass of AxA%)/100] + [(Mass of BxB%)/100] + [(Mass of CxC%)/100]
Atomic Mass = [(23.99x78.99)/100] + [(24.99x 10)/100] + [(25.98x11.01)/100]
Atomic number = 18.950 + 2.499 + 2.860
Atomic mass of Mg = 24.309
