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netineya [11]
4 years ago
15

Calculate Delta E when 33.0 g of carbon dioxide sublimes at 77.0 K and 1 ATM

Chemistry
1 answer:
bagirrra123 [75]4 years ago
3 0
To solve this problem, we establish the general energy balance:

ΔE = ΔU + ΔKE + ΔPE 
ΔE = Q + W

Q + W = ΔU + ΔKE + ΔPE

In this case, ΔKE and ΔPE are both zero or negligible.

Given:

m = 33.0 grams of CO2
Tsub = 77 K
P = 1 atm

ΔE = Q + W
ΔE = mCpΔT + ΔPV

solve for mCpΔT, find the value of Cp for CO2, then solve for Q. Next, solve for W using the ideal gas law. Add the two values and that will be the value of the delta E. 
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8 0
3 years ago
A particular laser consumes 130.0 Watts of electrical power and produces a stream of 2.67×1019 1017 nm photons per second.
solniwko [45]

The missing question is:

<em>What is the percent efficiency of the laser in converting electrical power to light?</em>

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

A particular laser consumes 130.0 Watt (P) of electrical power. The energy input (Ei) in 1 second (t) is:

Ei = P \times t = 130.0 J/s \times 1 s = 130.0 J

The laser produced photons with a wavelength (λ) of 1017 nm. We can calculate the energy (E) of each photon using the Planck-Einstein's relation.

E = \frac{h \times c }{\lambda }

where,

  • h: Planck's constant
  • c: speed of light

E = \frac{h \times c }{\lambda } = \frac{6.63 \times 10^{-34}J.s  \times 3.00 \times 10^{8} m/s }{1017 \times 10^{-9} m }= 6.52 \times 10^{-20} J

The energy of 1 photon is 6.52 × 10⁻²⁰ J. The energy of 2.67 × 10¹⁹ photons (Energy output = Eo) is:

\frac{6.52 \times 10^{-20} J}{photon} \times 2.67 \times 10^{19} photon = 1.74 J

The percent efficiency of the laser is the ratio of the energy output to the energy input, times 100.

Ef = \frac{Eo}{Ei} \times 100\% = \frac{1.74J}{130.0J} \times 100\% = 1.34\%

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

You can learn more about lasers here: brainly.com/question/4869798

8 0
3 years ago
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Answer:

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5 0
3 years ago
Read 2 more answers
For the reaction 2H2 + O2 --&gt; 2H2O, how many grams of oxygen are needed to react 3 moles of hydrogen?
mr Goodwill [35]

Answer:

48 grams

Explanation:

The chemical equation for the reaction is the following:

2 H₂ + O₂ → 2 H₂O

That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:

MW(O₂) = 16 g/mol x 2 = 32 g/mol

mass of O₂ = 1 mol x 32 g/mol = 32 g

So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:

3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂

<em>Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.</em>

4 0
3 years ago
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