Answer:
See explanation below
Explanation:
First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.
This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.
When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.
The next step involves the substraction of the hydrogen of the molecule by the Chlorine. This will leave the alkane with a lone pair available for reaction.
The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.
The final step involves forming the remaining products with the remaining reagents there.
In the picture attached you have the mechanism and product for this reaction:
Decreasing Pressure. The combined gas law states that the pressure of a gas is inversely related to the volume and directly related to the temperature. If temperature is held constant, the equation is reduced to Boyle's law. <em>Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase.</em>
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<span>Answer: The 1 kg of water will reach the lowest temperature
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Both of the objects is water so their specific heat should be same. The heat removed from the 2kg of water which is 2 times of mass than the 1kg water. Since the heat removed, both of their temperatures will drops. But 2kg water temperature drops will be half of 1kg water.
Compounds are not possible to be separated by physical means
Answer:
1.2 × 10^22 atoms.
Explanation:
Firstly, cations refers to the positively charged atom in the ionic compound, which is Na+.
Given the mass of NaCl as 1.17g, the number of moles of NaCl can be calculated this:
Molar mass of NaCl = 23 + 35.5
= 58.5g/mol
Mole = mass/molar mass
Mole = 1.17/58.5
Mole = 0.02moles
Using Avagadro's number, 6.022 × 10^23 atoms of Na+ are in 1 mole of NaCl.
In 0.02 moles of NaCl, there are 0.02 × 6.022 × 10^23 of Na+
0.1204 × 10^23 atoms
1.2 × 10^22 atoms of Na+ (cation)
