Ca + Cl2 = CaCl₂
A synthesis <span>reaction.</span>
C. Accelerating should be the correct onw
Answer:
1.5055×10²⁴ molecules
Explanation:
From the question given above, the following data were obtained:
Number of mole CO₂ = 2.5 moles
Number of molecules CO₂ =?
The number of molecules present in 2.5 moles CO₂ can be obtained as:
From Avogadro's hypothesis,
1 mole of CO₂ = 6.022×10²³ molecules
Therefore,
2.5 mole of CO₂ = 2.5 × 6.022×10²³
2.5 mole of CO₂ = 1.5055×10²⁴ molecules
Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂
Answer:
See Explanation
Explanation:
Metallic bonds involve attraction between electrons and positively charged metal ions. The metals are ionized and electrons form a sea of valence electrons. These loosely bound electrons surround the nuclei of the metals.
The presence of this sea of electrons explains the fact that metals conduct electricity and heat due to the free valence electrons.
Due to the nature of the bonding between metal atoms,metals are malleable and ductile.
Due to the strong electrostatic interaction between metal ions and electrons, the metallic bond is very strong and is very difficult to break thereby accounting for the greater strength of metals as the size of the metallic ion decreases.
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)
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
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