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2 years ago

K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s) net ionic equation

Chemistry

1 answer:

Oliga [24]2 years ago

3 0

Answer:

SO4 (aq) + Sr (aq) >> SrSO4 (s)

Explanation:

K+1 SO4-2 + Sr +2 I2- >> sK +1

(you dont separate and write down the ions for Sr SO4 because its a solid)

Cancel out the same ions from both sides and write down what your left with.

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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

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