<h3>
Answer:</h3>
70.906 g
<h3>
Explanation:</h3>
We are given;
- Atoms of Chlorine = 1.2 × 10^24 atoms
We are required to calculate the mass of Chlorine
- We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
- That is , 1 mole of an element = 6.022 × 10^23 atoms
- Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms
But since Chlorine gas is a molecule;
- 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms
But, molar mass of Chlorine gas = 70.906 g/mol
Then;
70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms
= 1.20 × 10^24 atoms
Thus;
For 1.2 × 10^24 atoms ;
= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)
<h3>= 70.906 g </h3>
Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g
=
4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3 is the reaction for preparation of silicon by the reduction of K₂SiF6 with Al.
AlF3xH2O-based inorganic compounds are referred to as aluminium fluoride. They are all solids without colour. Aluminium fluoride is a crystalline (sand-like), odourless, white, or colourless powder. In addition to being used to make aluminium, it also functions as a flux in welding processes and in ceramic glazes and enamels.
Silicon (Si) is created by reducing potassium silicofluoride with aluminium as the reducing agent (K2SIF6). While K2SiF6 is reduced to Si in this equation, aluminium is oxidised to aluminium fluoride. As a result, the balanced equation describing aluminum's reduction of K2SiF6 to silicon non-metal is as follows: 4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3
Learn more about aluminium fluoride here:
brainly.com/question/17131529
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<span>Since,
1000 grams of water = 1000 mL of water</span><span>
So,
At any of the given temperature:
</span>1000 mL = 10 x 100 mL
<span>
moles of NH4Cl = 53.5/53.49
= 1.0 m
= 1.0 mol/Kg
Delta T = 2 x 1.86 x 1.0
= 3.72 c
= - 3.72 °C</span>
Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
