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jolli1 [7]
4 years ago
8

A 12.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter with a heat capacity of 5.65 kJ/°C. Using the information be

low, determine the final temperature of the calorimeter if the initial temperature is 25.0°C. The molar mass of ethanol is 46.07 g/mol.
delta H rxn = -1235

C2H5OH + 3O2 -> 2CO2 +3H2O
Chemistry
2 answers:
Yanka [14]4 years ago
8 0
To answer the question above, we need to c<span>onvert 12.8 g to moles by dividing by 46.07 first.</span>

<span>For every mole you burn, you get 1235 kJ of heat. So multiply your number of moles by 1235. It'll be something in the neighborhood of 500. </span>

<span>Take your kJ that you calculated and divide by 5.65 to get the number of degrees that your calorimeter goes up. Add that to 25.

I hope my answer helped you</span>
pochemuha4 years ago
3 0

The final temperature of the calorimeter is 85.71 degree C.

Heat of the reaction is -1235 kJ/mol

Heat discharged at the time of reaction is 1235 kJ/mol

The moles of ethanol are calculated by using the formula mass of ethanol / molar mass of ethanol

= 12.8 g / 46 g/mol = 0.278 moles

Thus, the no. of moles of ethanol is 0.278 moles

The heat released when ethanol is combusted is:

0.278 moles × 1235 kJ/mol = 343 kJ

The final temperature is determined as,

343 kJ = (heat capacity) (temperature difference)

343 kJ = 5.65 (T - 25)

T-25 = 343 / 5.65

T-25 = 60.71

T = 85.71 degree C

Thus, the final temperature is 85.71 degree C.

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Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene
kenny6666 [7]

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}, 2-bromo-<em>2</em>-methylbutane;
  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}, 2-bromo-<em>1</em>-methylbutane.

It is expected that \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3} would end up being the dominant product.

Explanation

Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions \text{H}^{+} and negatively-charged bromide ions \text{Br}^{-}.

\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbo<em>cation</em> with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3};
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}.

The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has <em>three</em> carbon-carbon single bonds after the addition of the proton \text{H}^{+} as opposed to <em>two</em> as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.  

Both carbocations would then combine with bromide ions to produce a neutral halocarbon.  

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}

The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-<em>2</em>-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.

3 0
3 years ago
What is the solution to the problem expressed to the correct number of significant figures?
Fed [463]

Answer:

(102 900 ÷ 12) + (170 × 1.27) = 8800

Step 1. Evaluate the expressions inside the parentheses (PEMDAS)

102 900 ÷ 12 = 8575

170 × 1.27 = 215.9

In multiplication and division problems, your answer can have no more significant figures than the number with the fewest significant figures.

Thus, the underlined digits are not significant, but we keep them in our calculator to avoid roundoff error.

Step 2. Do the addition (PEMDAS).

  8575

+    215.9

= 8790.9

Everything that you add to an insignificant digit gives an insignificant digit as an answer.

Thus, the underlined digits are not significant.

We must drop them and round up the answer to 8800.

Explanation:

8 0
3 years ago
How many grams of NaF form when .508 mol of HF react with excess Na2SiO3?
jonny [76]
2HF + Na2SiO3 ---->  2NaF +  H2SiO3   is the balanced equation.

2 moles of HF  form 2 moles of NaF

therefore 0.508 Mol  HF forms 0.508 mol of NaF

In grams this is 0.508( rel atomic mass of Na + RAM of F)
                        = 0.508 (22.99+19)
                        =  21.33 grams 

8 0
4 years ago
I can model and explain how conduction transfers energy at a molecular level
astraxan [27]

Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more. These molecules then bump into nearby particles and transfer some of their energy to them. This then continues and passes the energy from the hot end down to the colder end of the substance.

Explanation:

3 0
3 years ago
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The thermometer in the picture reads, "19.9 degrees Celsius".
GalinKa [24]

Answer:

Sorry but i can't help

Explanation:

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6 0
3 years ago
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