Answer:
53.2 mL
Explanation:
As this problem deals with the neutralization of a strong monoprotic acid and a strong base, we can solve it by using the following formula:
Where in this case:
We <u>input the data</u>:
- 0.230 M * 74.0 mL = 0.320 M * Vb
And <u>solve for Vb</u>:
An atom that undergoes radioactive decay and has a large nucleus most likely contains....<span>- More protons than electrons.</span>
Answer:
the volume of CHCI3 = 7.87 ml
the volume of CHBr3 = 12.13 ml
Explanation:
From the given information:
We all know that 1 g/cm^3 = 1 g/ml
The density of boron = 2.34 g/ml
The Volume of the liquid mixture = 20 ml
Recall that:
Density = mass/volume
Mass = Density × Volume
Mass = 2.34 g/ml × 20 ml
Mass = 46.8 g
Suppose the volume of CHCI3 be Y and the Volume of CHBr3 be 20 - Y
Then :
Y (1.492) + 20-Y(2.890) = 46.8
1.492Y + 57.8 - 2.890Y = 46.8
- 1.398 Y = -11
Y = -11/ - 1.398
Y = 7.87 ml
Therefore, the volume of CHCI3 7.87 ml
the volume of CHBr3 = 20 - Y
= 20 - 7.87
= 12.13 ml
Answer:
Post it on another page I can try to help.