Part b which of the isoelectronic pairs you determined above has the same electron configuration?

How many moles of Cobalt are in 27.4 grams of cobalt

xeze [42]

0.4649331785818406 is what 27.4 grams is converted to! You're welcome!! :)

5 0

2 years ago

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How many total ions are present in 347g of cacl2?

vladimir1956 [14]

In 1 mole of $CaCl_{2}$ , there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of $Cl^{-1}$ .

$CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}$

Molar mass of $CaCl_{2}$ is 110.98 g/mol. Calculating number of moles from given mass as follows:

$n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol$

Thus, number of moles of ions will be $3\times 3.12 mol=9.38 mol$ .

Since, 1 mole of any substance has $6.023\times 10^{23}$ units of that substance where $6.023\times 10^{23}$ is Avogadro's number.

Thus, 9.38 mol of ions will have $9.38\times 6.023\times 10^{23}=5.65\times 10^{24}$ number of ions.

Therefore, total number of ions in 347 g of $CaCl_{2}$ is $5.65\times 10^{24}$ .

8 0

2 years ago

A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0

2 years ago

On the graph, indicate the distance that corresponds to the bond length of the N2 molecule by placing an X on the horizontal axi

Brut [27]

Answer:

Check the image in the file attached below.

Explanation:

The objective of this question is to use the diagram provided to indicate the distance that corresponds to the bond length of the N2 molecule by placing X on the horizontal axis.

The bond length shows the distance of the bonds between the nuclei of the bonded atoms between the N2 molecule and O2 molecule. From the diagram, we can see that the bond length of O2 shows a greater tendency over the N2 molecule and it also have higher potential energy than the N2 molecule.

6 0

2 years ago

Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.

Lelu [443]

Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.

4 0

2 years ago