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2 years ago

79. Methane, the chief component of natural gas, burns in oxygen according to the equation

Chemistry

1 answer:

Andrews [41]2 years ago

6 0

Answer:

14 mol

Explanation:

CH4(g) + 2O2(g) -------» CO2(g) + 2H2O(g)

from reaction 1 mol 2 mol

given 7 mol x mol

x = (7*2)/1 = 14 mol

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What group of professionals specialize in designing, testing, and building the products of technology?

swat32

Answer:

Explanation:

4 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:

What is the percent efficiency of the laser in converting electrical power to light?

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

2 years ago

The ksp of copper(ii) carbonate, cuco3, is 1.4 × 10-10. calculate the molar solubility of this compound.

mafiozo [28]

As,
CuCO₃   ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² =  1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x =  1.18 × 10⁻⁵ mol/L × 123.526 g/mol

x = 1.45 × 10⁻³ g/L

3 0

2 years ago

Measurements include

Keith_Richards [23]

There’s lots of measurements. (m, kg, s, mol, cm, in, mm) etc

8 0

3 years ago

If 38.5 grams of potassium react with excess oxygen gas, how many grams of potassium oxide can be produced? 4K + O2 yields 2K2O

Lera25 [3.4K]

Answer:

46.40 g.

Explanation:

It is a stichiometric problem.

The balanced equation of the reaction: 4K + O₂ → 2K₂O.

It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.

We should convert the mass of K (38.5 g) into moles using the relation:

n = mass / molar mass,

n = (38.5 g) / (39.098 g/mol) = 0.985 mole.

Using cross multiplication:

4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.

0.985 mole of K produces → ??? moles of K₂O.

∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.

Now, we can get the mass of K₂O:

∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.

6 0

2 years ago