Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:
A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.
Therefore the only choice that fits is A. 80 g of Br2.
Answer:
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is :
concentration of 
= 10%
10 g of in 100 ml solution
molar mass = 45.08 g/mol
number of moles = 10 / 45.08
= 0.222 mol
Molarity of 
= 2.22 M
number of moles of in 20 mL can be determined as:
Concentration of 
= 2.22 M
Similarly, The pKa Value of is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
the pH value at equivalence point is,
![pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]](https://tex.z-dn.net/?f=pH%3D%20%5Cfrac%7B1%7D%7B2%7DpKa%20-%20%5Cfrac%7B1%7D%7B2%7DpKb-%5Cfrac%7B1%7D%7B2%7Dlog%5BC%5D)
![pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]](https://tex.z-dn.net/?f=pH%20%3D%20%5Cfrac%7B14%7D%7B2%7D-%5Cfrac%7B3.25%7D%7B2%7D-%5Cfrac%7B1%7D%7B2%7Dlog%20%5B2.22%5D)
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
Ignition wires make it more accurate because it will cook it faster
stirrer would have the less results of fast
a sealed bomb may cook it fast but you would have to be careful and don't mess up
Answer:
v = 23.96 cm³
Explanation:
Given data:
Mass = 15.0 g
Density = 0.626 g/cm³
Volume = ?
Solution:
Formula:
D=m/v
D= density
m=mass
V=volume
Now we will put the values in formula:
d = m/v
v = m/d
v = 15 g / 0.626 g/cm³
v = 23.96 cm³
Answer:
48.67 seconds
Explanation:
From;
1/[A] = kt + 1/[A]o
[A] = concentration at time t
t= time taken
k= rate constant
[A]o = initial concentration
Since [A] =[A]o - 0.75[A]o
[A] = 0.056 M - 0.042 M
[A] = 0.014 M
1/0.014 = (1.1t) + 1/0.056
71.4 - 17.86 = 1.1t
53.54 = 1.1t
t= 53.54/1.1
t= 48.67 seconds
Hence,it takes 48.67 seconds to decompose.
