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2 years ago

How many moles of carbon dioxide will be produced from the complete combustion of 13.6 moles of butane?

Chemistry

1 answer:

Aleks04 [339]2 years ago

5 0

Answer:

54.4 mol

Explanation:

the equation for complete combustion of butane is

2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O

molar ratio of butane to CO₂ is 2:8

this means that for every 2 mol of butane that reacts with excess oxygen, 8 mol of CO₂ is produced

when 2 mol of C₄H₁₀ reacts - 8 mol of CO₂ is produced

therefore when 13.6 mol of C₄H₁₀ reacts - 8/2 x 13.6 mol = 54.4 mol of CO₂ is produced

therefore 54.4 mol of CO₂ is produced

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Whats the answer? please help me

dybincka [34]

Answer C
I could be wrong
Explanation

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3 years ago

The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching it

ollegr [7]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

$5 \times 10^{26} atoms = 1 [tex]m^{3}$

2 atoms = $\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms$

= $4 \times 10^{-27} m^{3}$

Formula for volume of a cube is $a^{3}$ . Therefore,

Volume of the cube = $4 \times 10^{-27} m^{3}$

As lattice constant (a) = $(4 \times 10^{-27} m^{3})^{\frac{1}{3}}$

= $1.59 \times 10^{-9} m$

Therefore, the value of lattice constant is $1.59 \times 10^{-9} m$ .

And, for bcc unit cell the value of radius is as follows.

r = $\frac{\sqrt{3}}{4}a$

Hence, effective radius of the atom is calculated as follows.

r = $\frac{\sqrt{3}}{4}a$

= $\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m$

= $6.9 \times 10^{-10} m$

Hence, the value of effective radius of the atom is $6.9 \times 10^{-10} m$ .

3 0

3 years ago

An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize

oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

8 0

3 years ago

Which of the following is not equal 225 g?

vekshin1

G to dkg is 1 place to the right, move the decimal point 1 place to the left
<span>225 g = 22.5 dkg
C is the answer</span>

3 0

3 years ago

Read 2 more answers

What is the answer? <br><br> NO LINKS!!

jek_recluse [69]

Liquid? maybe, its really inbetween if you get what i mean

5 0

2 years ago