What is the molecular geometry of carbon tetrachloride (CC1,)?

Chemistry

1 answer:

dangina [55]3 years ago

6 0

Answer:

tetrahedral

As carbon has no 4 valence electrons that are occupied by 4 chlorine atoms, we get 0 lone pair in carbon and sp3 hybridisation which leads to tetrahedral shape.

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What is the molality of an aqueous solution that contains 29.5 g of glucose (C6H12O6) dissolved in 950 g of water (H2O)?

Gnesinka [82]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
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3 years ago

A sample of vinegar has a hydronium concentration of 1.6 × 10-3 M. What is the pH of the vinegar?

loris [4]

PH is defined as the negative log of Hydronium Ion concentration.

So, in order to find the pH of vinegar, we find the negative log of its hydronium ion concentration.

$pH=-log[ H_{3}O^{+}] \\ \\ 
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So, the pH of given vinegar solution will be 2.8.

Therefore, the answer to this question is option B

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Each molecule of table sugar, C12H22O11, contains

Delicious77 [7]

D, because C12 means there's 12 atoms of carbon.

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During a hurricane, what effect can the ocean have on the beach?

Katen [24]

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A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

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3 years ago