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egoroff_w [7]
3 years ago
12

How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?

Chemistry
1 answer:
xenn [34]3 years ago
4 0

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

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Look at this compound, why does magnesium has dative and covalent bonds with nitrogen? ...?
HACTEHA [7]
<span>The solid lines between N and Mg are actually ionic bonds. N has 5 valence electrons (2 of which are paired). Of the 3 that are unpaired, 2 are part of covalent bonds with adjacent carbon atoms. N accepts an extra electron to complete its octet, but gets a formal charge of -1. This allows for formation of an ionic bond with Mg, which is +2. Two of these charged N atoms therefore neutralize the charge of the central Mg. As for the coordinate (dative) covalent bonds, Mg has empty orbitals - the ionic bonds with the charged N atoms give it only 4/8 possible valence electrons.


The other two N atoms (dotted lines) have a formal charge of 0 since they form three covalent bonds with adjacent carbon atoms, but they still have a lone pair. Therefore, just to improve stability, each of these N atoms can "donate" its lone pair to Mg in order to complete its octet.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!



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4 0
4 years ago
The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

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1 year ago
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Answer:

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Explanation:

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