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.
To know more about moles, visit the below link:
brainly.com/question/15303663
It would take 8 antacid tablets to produce 120 mL of CO2 gas.
Answer:
Making oxygen
Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:
hydrogen peroxide → water + oxygen
2H2O2(aq) → 2H2O(l) + O2(g)
The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.
Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.
To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.
Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.
Answer: -105 kJ
Explanation:-
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E%28reactant%29%5D-%5Csum%20%5Bn%5Ctimes%20B.E%28product%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN_2%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH_2%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%20B.E_%7BNH_3%7D%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN%5Cequiv%20N%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH-H%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%203%5Ctimes%20B.E_%7BN-H%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20945%29%2B%283%5Ctimes%20432%29%5D-%5B%282%5Ctimes%203%5Ctimes%20391%29%5D)
Therefore, the enthalpy change for this reaction is, -105 kJ
They are good conductors of heat and electricity.
They are solid at room tempature
They have a high melting point
