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
3.6*10^20 atoms* (1 mol/ (6.02*10^23 atoms))= 6.0*10^(-4) mol.
6.0*10^(-4) mol* (28.1 g/ 1 mol)= 1.7*10^(-2) g.
Note that the units cancel out so you get the answer.
The final answer is 1.7*10^(-2) g Si.
Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions. The “ingredients” of a reaction are called reactants, and the end results are called products.
Carbon electrodes are changed frequently in electrolysis as the carbon in the carbon electrodes burns to give out Carbon-di-oxide.
Carbon electrodes are used in electrolysis to facilitate the transfer of electrons between the anode and the cathode. The carbon electrode is usually in the form of a rod or plate and is electrically connected to the anode.
The carbon electrode may also be used to help reduce the overall resistance of the electrolytic cell.
Carbon electrodes are used in electrolysis because they are chemically inert and have a high electrical conductivity.
However, carbon electrodes must be changed frequently because they become corroded and lose their electrical conductivity over time.
This is because carbon burns out in the Carbon electrodes to give out Carbon-di-oxide. That is why the carbon anodes should be changed frequently.
To know more about electrolysis, click below:
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Answer: 2.5g
Explanation:
5 % sucrose solution means that 5 % of the weight of the solution is sucrose.
If 1 liter of water weighs 1000 grams. To prepare 5% sucrose solution 5/100 x 1000 = 50 grams. Since 1 liter equals 1000ml, thus a 5 % solution has 50 grams of solute dissolved in one liter.
To prepare 5% sucrose solution in 50mls
=5/100 x 50
= 0.05 x 50
= 2.5g
Therefore to prepare 5% sucrose solution in 50mls we dissolve 2.5g of sucrose in 50ml of water
