No of Neutrons =Mass No-No of protons
- Hence it has atomic no 38
The element is Stroncium(Sr)
The reactions are in order which includes combustion reaction, Hydration reaction, oxidation reaction, and displacement reaction.
a) A combustion reaction is a chemical reaction between a fuel and an oxidant where heat is released. The combustion reaction example is given below. It is a balanced chemical reaction.
2C₃H₆(g) + 9O₂(g) --------> 6CO₂(g) + 6H₂O(g)
b. A hydration reaction is a chemical reaction in which a molecule of water is added to another molecule. Here Aluminum oxide is added to water to form aluminum hydroxide.
4Al₂O3(s) + 6H₂O(l)------> 2Al(OH)3(s)
c. When a metal reacts with oxygen, the metal forms an oxide. Oxide is a compound of metal and oxygen. Here lithium metal reacts with oxygen to form lithium oxide.
2Li(s) + O₂(g)-----> Li₂O(s)
d. A displacement reaction is one in which a more reactive element displaces a less reactive element from a compound. Here Zinc is more reactive than silver, so silver was displaced to form Zinc Nitrate.
Zn(s) + 2AgNO₃(aq) -----> 2Ag(s) + Zn(NO₃)₂(aq)
To know more about reactions, click below:
brainly.com/question/11231920
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Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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The balanced chemical
reaction will be:
CH4 + 2O2 → CO2 + 2H2O
We are given the amount of carbon dioxide to produce from the reaction.
This will be our starting point.
560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) (
22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>
Answer:
1. C = 0.73 M.
2. pH = 0.14
Explanation:
The reaction is the following:
HCl + NH₃ ⇄ NH₄⁺Cl⁻
From the titration, we can find the number of moles of HCl that were neutralized by the ammonia.
Where "a" is for acid and "b" is for base.
The number of moles is:
Where "C" is for concentration and "V" for volume.
Hence the initial concentration of the acid is 0.73 M.
The original pH of the acid is given by:
![pH = -log([H^{+}])](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH%5E%7B%2B%7D%5D%29%20)
Therefore, the original pH of the acid is 0.14.
I hope it helps you!
