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geniusboy [140]

2 years ago

10

Butane is burned as a lighter fluid in disposable lighters. Write a balanced equation for the complete oxidation reaction that o

ccurs when butane (C4H10) burns in air. Use the smallest possible integer coefficients.

1 answer:

larisa [96]2 years ago

7 0

Answer:

The balanced equation for the complete oxidation reaction uses 2 moles of butane and 13 moles of Oxygen.

Explanation:

The oxidation of butane is shown in the following picture. As it is a complete combustion, the products of this reaction are carbon dioxide and water.

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Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

a)

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

b)

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

3 years ago

Read 2 more answers

Which statement is true for one molecule of sulfur trioxide?A.(There are three atoms of sulfur and one atom of oxygen.B.(There a

ki77a [65]

The answer would be D. This is because sulfur is on it's own, meaning one. while tri is a prefix for three so there are three oxygen atoms.

5 0

3 years ago

Read 2 more answers

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

What is the volume, in mL, of 0.50 moles of ammonia (NH3) at STP?

blagie [28]

For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= ?
n = 0.5 moleT= 273.15 K (at STP)
p= 101.325 kPa (at STP)
R is universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:
V=nxRxT/p
V=0.5x8.314x273.15/101.325= 11.2 L = 11200 mL

Answer: D.

5 0

2 years ago

Hydrochloric acid (HCI) is created by a chemical reaction between hydrogen (H) and chlorine (CI). Which substance is

Mashutka [201]

The reaction and product is

H2+Cl2 = 2HCl

8 0

3 years ago