Well the the answer is 70.8c but if you round it up it is 71c which I choice and got it correct so the answer is 71c
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
Based on the information given, it should be noted that the ground-state electron configuration of carbon is 1s2 2s2 2p2.
<h3>
What is an electron?</h3>
Electrons are simply the subatomic particles which orbit the nucleus of an atom.
The arrangement of electrons in the atomic orbitals of an atom is known as the electron configuration. This can be determined by using a periodic table.
It should be noted that carbon is the sixth element with a total of 6 electrons in the periodic table. Thus, the atomic number Z = 6.
In conclusion, the ground-state electron configuration of carbon is 1s2 2s2 2p2.
Learn more about carbon on:
brainly.com/question/105003
Answer:2C12
Explanation:because if u divide the the equation by 2 u well get the same answer you can multiply but that will take longer with the invisible zeros and stuff like that but the answer is 2c12 and on the real FSA u can pull out ur phone and cheat just make sure the teacher is not watching just playing with you DONT DO THAT YOU WILL GET IN SERIOUS TROBLE KIDS DO NOT DO THIS KN THE REAL DAY OF THE FSA PLEASE DONT DO IT YOU WILL GET IN SERIOUS TROUBLE WELL THATS IT FOR TODAY SO HAVE A GREAT DAY THIS IS TO PUT PRESSURE ON YOU
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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