The answer:
<span>The equation of its dissolution in water is: AgNO3 → Ag + (aq) + NO3- (aq)
and </span>AgNO3 → Ag + (aq) + NO3- (aq)
1 mol 1mol 1mol
? -------- 0.854mo
so for finding the value, it is sufficients to complute 1 x 0.854 mol =0.854 mol
so, 0.854 mol is required for the reaction to form 0.854 mol of Ag
Answer:
0.019 moles of M2CO3
Explanation:
M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)
From the equation above;
1 mol of M2CO3 reacts to produce 1 mol of BaCO3
Mass of BaCO3 formed = 3.7g
Molar mass of BaCO3 = 197.34g/mol
Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol
Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,
1 = 1
x = 0.019
x = 0.019 moles of M2CO3
Answer:
The atomic mass of second isotope is 7.016
Explanation:
Given data:
Average Atomic mass of lithium = 6.941 amu
Atomic mass of first isotope = 6.015 amu
Relative abundance of first isotope = 7.49%
Abundance of second isotope = ?
Atomic mass of other isotope = ?
Solution:
Total abundance = 100%
100 - 7.49 = 92.51%
percentage abundance of second isotope = 92.51%
Now we will calculate the mass if second isotope.
Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
6.941 = (6.015×7.49)+(x×92.51) /100
6.941 = 45.05235 + (x92.51) / 100
6.941×100 = 45.05235 + (x92.51)
694.1 - 45.05235 = (x92.51)
649.04765 = x
92.51
x = 485.583 /92.51
x = 7.016
The atomic mass of second isotope is 7.016
Answer:
PV=nRT
n = PV/RT
n = m/Mm
m/Mm = PV/RT
m = MmPV/RT
T in kelvin = T Celsius + 273.15 = 293.15 K
m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)
mass in grams = 82.8 grams
Explanation:
Ideal gases formula is PV=nRT, where:
P is the pressure (1.39 atm in this case)
V is the volume (55.0 L in this case)
R is the gas constant (0.08206 L.atm/K.mole)
T is the temperature (20.0C) should be converted to Kelvin
all the unit should correspond to the one in the R.
we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.
we found the mass to be 82.8 grams
It matches the universal pH indicator and is indicating the proper pH