Answer:
6.17 g/cm³
Explanation:
Data given:
one side of cube = 0.53 cm
mass of the cube is 0.92 g
density of the cube = ?
Solution:
First we will calculate for volume the cube
As we know all the sides or edges of a cube are equal so volume equation will be
So,
V = length x width x height
V = e³
as on side = 0.53 cm
then
V = (0.53 cm)³
V = 0.149 cm³
Now we will calculate density of cube
To calculate density, formula will be used
d = m/v . . . . . (1)
where
d = density
m = mass
v = volume
put values in above formula 1
d = 0.92 g / 0.149 cm³
d = 6.17 g/cm³
so. the density of cube = 6.17 g/cm³
Answer:
see explanation
Explanation:
Write the balanced COMPLETE ionic equation for the reaction when Na₂CO₃ and AgNO₃ are mixed in aqueous solution. If no reaction occurs, simply write only NR.
Ag (+1) + NO3(-1) + 2 Na(+1) + Co3 (-2)--> Ag2CO3 (s) + 2 Na (+1) + 2NO3(-1)
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
One separation technique to be used is the paper chromatography. This works by separating the components of the mixture through the difference of their concentrations. There is a stationary phase and the mobile phase, which flows through the stationary phase. The components travel at different rates and is usually signified by the colors. If more than one color would appear, that means that the dye is a mixture.
