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3 years ago

2c4H10 +13O2 —>8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10

Chemistry

1 answer:

Elena-2011 [213]3 years ago

5 0

Answer:

$\large\boxed{\text{28 mol of O$_{2}$}}$

Explanation:

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol: 4.3

13 mol of O₂ react with 2 mol of 2C₄H₁₀

$\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}$

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Compare and contrast chemical potential energy and elastic potential energy

nikdorinn [45]

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Elastic Potential is released when the molecules in the material are allowed to go back to there original form, and is released mainly as kinetic

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3 years ago

The reaction a(g)⇌b(g) has an equilibrium constant of 5.8 and under certain conditions has q = 336. part a what can you conclude

Vika [28.1K]

Answer:

The answer is "As $Q=336$ , at high-temperature $\Delta G_{rxn}>0$ and When $K>1,$ $\Delta G^{\circ}_{rxn}>0$ ."

Explanation:

The equation for the reaction is:

$A(g) \leftrightharpoons B(g)$

$K=5.8\\\\Q=336$

At equilibrium,

$\Delta G^{\circ}_{rxn}>0$ $=-RT \ln \ K$

When k=5.8(>1), the value of $\ln k$ would be positive

So, $\Delta G^{\circ}_{rxn}$ is negative (< 0)

So if K > l, $\Delta G^{\circ}_{rxn}$

If the reaction is not in equilibrium so the equation is :

$\Delta G_{rxn}>0$ = $\Delta G^{\circ}_{rxn}$ $+RT \ln Q$

Substituting the expression:

$\Delta G_{rxn}>0$ $= (-RT \ln K) + RT \ln Q$

$= RT(\ln Q- \ln K)\\= RT(\ln (336)-\ln (5.8))\\= RT(4.06)$

It is the positive value for all temperatures.

So, As Q = 336, at the high temperature $\Delta G_{rxn}>0$ .

5 0

3 years ago

What is the molarity of an aqueous solution that contains 78g of C6H12O6 dissolved in 2500 mL of solution?

dusya [7]

Answer:

$\boxed {\boxed {\sf molarity = 0.17 \ M \ C_6H_12O_6}}$

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

$molarity = \frac {moles}{liters}$

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.

1. Gram to Moles

We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.

C: 12.011 g/mol
H: 1.008 g/mol
O: 15.999 g/mol

Next, look at the formula and note the subscripts. This tells us the number of atoms in 1 molecule. We multiply the molar mass of each element by its subscript.

6(12.011)+12(1.008)+6(15.999)=180.156 g/mol

Use this number as a ratio.

$\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}$