Calculate the mass of butane needed to produce 96.7 g of carbon dioxide

Answers

Answer 1

31.9 grams butane needed to produce 96.7 grams CO2

Related Questions

A metal M is converted to the metal sulfate M2(SO4)3. then a soultion of the metal sulfate was treated with calcium chloride to give calcium sulfate perciptate.If 1.2 g of the metal gave 5.451 g calcium sulfate what is the identity of metal M?
M2(SO4)3(aq)+3CaCl2(aq) ----->
2MCl3(aq)+3CaSO4(s)

Answers

1) From the chemical equation you obtaing the theoretical molar ratios:

2 mol of M : 3 mol of CaSO4

2) Convert 5.451 g of CaSO4 into number of moles

number of moles = mass in grams / molar mass

molar mass 0f CaSO4 = 136.14g/mol

=> number of moles of CaSO4 = 5.451 g / 136.14 g/mol = 0.0400 mol

3) Make a proportion with the unknown, 0.0400 mol and the theoretical molar ratio:

x / 0.0400 mol of CaSO4 = 2 mol of M / 3 mol of CaSO4

=> x = 0.400 * 2 / 3 mol of M = 0.02667 mol of M

4) Use the mass of 1.2 g and the number of moles to obtain the molar mass of M

molar mass = mass in grams / number of moles

=> molar mass = 1.2 g / 0.02667 moles = 45 g/mol

5) Use a periodic table to find the molar masses of different metals.

Scandium atomic mass is 45 g/mol, so the metal is Scandium.

Answer: Scandium

A certain shade of blue has a frequency of 7.03 × 1014 Hz. What is the energy of exactly one photon of this light?

Answers

4.66 x 10^-19 Joules The energy of a photon is E=hf Where E = Energy h = Planck's constant (6.62607004Ă—10â’34 J s) f = frequency of photon. So plug in the known values E = 6.62607004Ă—10â’34 J s * 7.03 x 10^14 1/s E = 4.65812724x10^-19 J Since we only have 3 significant figures, the rounded result is E = 4.66x10^-19 J

The energy of one photon of light with a frequency of 7.03 × 1014 Hz is approximately 4.65618 × 10-19 Joules. This is obtained using Planck's equation, E = hf, where E is the energy, h is Planck's constant and f is the frequency of the light.

Explanation:

The subject of this question is Physics. The student wants to know the energy of exactly one photon of light that has a frequency of 7.03 × 1014 Hz. We can use Planck's equation which relates the energy of photon, its frequency and Planck's constant (h = 6.626 × 10-34 J.s).

So, the energy (E) of the photon is given by the formula E = hf. Substituting the frequency (f = 7.03 × 1014 Hz) and Planck's constant into the formula gives E = (6.626 × 10-34 J.s)(7.03 × 1014 Hz) which gives approximately 4.65618 × 10-19 Joules. This is the energy of one photon of light with a frequency of 7.03 × 1014 Hz.

Learn more about Photon here:

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