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statuscvo [17]
3 years ago
6

The bond length of the chlorine molecule, Cl2, is 199 pm. On the other hand, a pure carbon structure, such as diamond, has C-C b

ond lengths of 154 pm. What is the bond length of a C-Cl bond?
Chemistry
1 answer:
Finger [1]3 years ago
4 0

Answer:

176.5 picometer is the bond length of a C-Cl bond.

Explanation:

Bond length of chlorine molecule = 199pm

Radius of the chlorine atom = r_{Cl}=\frac{199 pm}{2}

Bond length of carbon molecule = 154 pm

Radius of the carbon atom = r_c=\frac{154 pm}{2}

Bond length of C-Cl =

r_c+ r_{Cl}=\frac{154 pm}{2}+\frac{199 pm}{2}=176.5 pm

176.5 picometer is the bond length of a C-Cl bond.

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Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
2 years ago
Why is the? The second ionization energy of sodium is about three times greater than the second ionization energy of magnesium
Scorpion4ik [409]

Answer:

~Na+1 is already in the preferred form. Because of this, the second ionization energy of sodium is higher than normal. Mg+1 loses an electron to form s2 p6 .

3 0
2 years ago
When a teacher dilutes 50 mL of 2.0 M NaOH to 0.50 M, what volume of NaOH results?
JulijaS [17]

Answer:

= 200 mL

Explanation:

Using the dilution formula;

M1V1 = M2V2 ;

Where, M1 is the concentration before dilution, V2 is the volume before dilution, while M2 is the concentration after dilution and V2 is the volume after dilution.

M1 = 2.0 M

V1 = 50 mL

M2 = 0.50 M

V2 = ?

V2 = M1V1/M2

     = ( 2.0 × 50 )/ 0.5

     = 200 mL

Therefore, the volume after dilution will be, 200 mL

6 0
3 years ago
g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) +
iren2701 [21]

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Mg: 1 mole
  • HCl: 2 moles
  • MgCl₂: 1 mole
  • H₂: 1 mole

Being the molar mass of each compound:

  • Mg: 24.3 g/mole
  • HCl: 36.45 g/mole
  • MgCl₂: 95.2 g/mole
  • H₂: 2 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Mg: 1 mole* 24.3 g/mole= 24.3 g
  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • MgCl₂: 1 mole* 95.2 g/mole= 95.2 g
  • H₂: 1 mole* 2 g/mole= 2 g

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

Molarity=\frac{number of moles of solute}{volume}

in units \frac{moles}{liter}

then, the number of moles of HCl that react is:

6 M=\frac{number of moles of HCl}{0.01 L}

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, <u><em>Mg will be the limiting reagent.</em></u>

7 0
3 years ago
"Infant Joy" is a conversation between __.
ElenaW [278]
The answer to  this is C.
6 0
2 years ago
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