Answer:
Moles of BCl₃ needed = 0.089 mol
Explanation:
Given data:
Moles of BCl₃ needed = ?
Mass of HCl produced = 10.0 g
Solution:
Chemical equation:
BCl₃ + 3H₂O → 3HCl + B(OH)₃
Number of moles of HCl:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 36.46 g/mol
Number of moles = 0.27 mol
Now we will compare the moles of HCl with BCl₃.
HCl : BCl₃
3 : 1
0.27 : 1/3×0.27 = 0.089 mol
Because there are many numbers of the human measurement . we must manage the uncertainly doing calculations because we can know what we are calculating.
It is b because I'm just going around the house with me to the mall to get u at the moment I don't have any plans tonight or naw
Answer:
mass of U-235 = 15.9 g (3 sig. figures)
Explanation:
1 atom can produce -------------------------> 3.20 x 10^-11 J energy
x atoms can produce ----------------------> 1.30 x 10^12 J energy
x = 1.30 x 10^12 / 3.20 x 10^-11
x = 4.06 x 10^22 atoms
1 mol ----------------------> 6.023 x 10^23 atoms
y mol ----------------------> 4.06 x 10^22 atoms
y = 0.0675 moles
mass of U-235 = 0.0675 x 235 = 15.8625
mass of U-235 = 15.9 g (3 sig. figures)
The temperature that would the volume of a gas be 0.550l if it had a volume of 0.432 L at -20.0 c is calculated using the Charles law formula
that is v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l
by making T1 the subject of the formula T1= V1T2/V2
T1= (0.55lL x253)/ 0.432 l = 322.11 K or 322.11-273 = 49.11 C
