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lianna [129]
3 years ago
5

If the heat of combustion for a specific compound is −1160.0 kJ/mol and its molar mass is 86.47 g/mol, how many grams of this co

mpound must you burn to release 541.20 kJ of heat?
Chemistry
1 answer:
notsponge [240]3 years ago
8 0

Answer:

40.34 g

Explanation:

First, we divide the heat to release by the heat of combustion to obtain the required moles of compound:

541.20 kJ/(1160.00 kJ/mol) = 0.4665 mol

So, we have to burn approximately 0.47 mol of the compound. We convert the moles to mass in grams by using the molar mass:

mass = molar mass x moles = 86.47 g/mol x 0.4665 mol = 40.34 g

Therefore, you must burn 40.34 grams of the compound to release 541.20 kJ of heat.

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Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp
notka56 [123]

Answer:

0.00 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.  

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ            30.07      32.00  

              2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g:      1.50          11.

2. Calculate the moles of each reactant  

\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

\text{Moles of CO}_{2} =  \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

8 0
3 years ago
What does the stonefish eat
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Croat or stones stones
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4 years ago
A molecule of ethyl alcohol is converted to acetaldehyde in one’s body by zero order kinetics. If the concentration of alcohol i
Oliga [24]

Answer:

(1) 0.0016 mol/L

Explanation:

Let the concentration of alcohol after 3.5 hours be y M

The reaction follows a first-order

Rate = ky^0 = change in concentration/time

k = 6.4×10^-5 mol/L.min

Initial concentration = 0.015 M

Concentration after 3.5 hours = y M

Time = 3.5 hours = 3.5×60 = 210 min

6.4×10^-5y^0 = 0.015-y/210

y^0 = 1

0.015-y = 6.4×10^-5 × 210

0.015-y = 0.01344

y = 0.015 - 0.01344 = 0.00156 = 0.0016 mol/L (to 4 decimal places)

5 0
3 years ago
Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break
KIM [24]

Answer:

a) Endothermic

b) T₂ = 53.1 ºC

Explanation:

a) We are told that when the ammonium nitrate dissolves in water the pack gets cold so the system is absorbing heat from the surroundings and by definition it  is an endothermic process.

b) Recall that the heat, Q, is given by the formula:

Q = mcΔT    where  m is the mass of water,

                                 c is the specific heat of water, and

                                 ΔT is the change in temperature

We can determine the value for Q since we are given the heat of solution for the ammonium nitrate. From there we can calculate ΔT and finally answer our question.

Molar mass NH₄NO₃ = 80.04 g/mol

moles NH₄NO₃ = 50.0 g/ 80.04 g/mol = 0.62 mol

Q = 25.4 kJ/mol x 0.62 mol = 15.87 kJ = 15.87 kJ x 1000 J = 1.59 x 10⁴ J

Q = mcΔT ⇒ ΔT = Q/mc

ΔT = 1.59 x 10⁴ J/ (135 g x 4.184 J/gºC ) = 28.1 ºC

T₂- T₁ = ΔT ⇒ T₂ = ΔT  + T₁  = 28.1 ºC +25.0 ºC = 53.1 ºC

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What role does DNA play in the production of a protein?
klemol [59]

A

Explanation:

The coiling up of DNA with the help of histone proteins - to what is called heterochromatic regions-  is part of gene regulation. This makes genes inaccessible to RNA polymerase that makes mRNA from the genes. When the genes are exposed by DNA unwinding, these genes are transcribed and the resulting mRNAs are translated by ribosomes into proteins.

The DNA never unwinds completely, but rather does so region by region, because if it does so it would become so long that it wouldn't fit in the nucleus or cell.

5 0
3 years ago
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