The moles of I₂ will form from the decomposition of 3.58g of NI₃ is 0.0136 moles.
<h3>How we calculate moles?</h3>
Moles of any substance will be calculated as:
n = W/M, where
W = required mass
M = molar mass
Given chemical reaction is:
2NI₃ → N₂ + 3I₂
Moles of 3.58g of NI₃ will be calculated as:
n = 3.58g / 394. 71 g/mol = 0.009 moles
From the stoichiometry of the solution, it is clear that:
2 moles of NI₃ = produce 3 moles of I₂
0.009 moles of NI₃ = produce 3/2×0.009=0.0136 moles of I₂
Hence, option (3) is correct i.e. 0.0136 moles.
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S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
The gas is ignited (I think) and combustion happens where the gasoline turns into gas (the state of being) and expands, pushing something and making the blades turn so
from stationary to explosive so potentioal to kenetic
<h2>Heptene formed is -</h2><h2>
</h2>
Explanation:
The two possibilities when the peroxide is not present
- + HBr →
In presence peroxide,
≡+ HBr →
- When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
- This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
- One mole of HBr adds to one mole of 1-heptane.
- The structure of heptene formed is -