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3 years ago

Please hurry!!! And help!!

Chemistry

1 answer:

Crazy boy [7]3 years ago

6 0

Answer:

Graphite, because it has atoms of same atomic and electrovalent electrons

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Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

When three pairs of electrons are shared in a molecule, what shape will the molecule be?

djyliett [7]

The correct answer is B. The shape of a molecule where three pairs of electrons are shared is a trigonal planar. This is characterized by one central atom and three atoms forming an equilateral triangle which is bonded to the central atom.

8 0

3 years ago

Read 2 more answers

What is the mass of 22.4 L of H2 at STP? (5 points)

cestrela7 [59]

11.2
Step-by-step-explanation

3 0

3 years ago

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CH3CH2OH + 302 → 2C02 + 3H2O

Colt1911 [192]

Answer:

Oxygen Gas

Explanation:

The balanced equation shows us the reactant ratio of the reaction.

This means that for every one mole of CH3CH2OH, we need 3 moles of O2 to react with it. Because we need more O2, (3x as much) than ethanol and we have the same given amount (1 mole of each), the oxygen will be the limiting reagent. (1 mole of oxygen would only require 1/3 moles of ethanol to react).

Hope this helped!

7 0

3 years ago

Helppppp asaapppppp plzzzzzz

Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

(0.75M) x (0.094L) = C2

(0.329L)

C2 = 0.214 M (Rounded)

When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.

7 0

2 years ago