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2 years ago

Consider the following equation: K + Br → KBr. How many grams of KBr can you make with 1.25g of K?

Chemistry

1 answer:

Dmitry [639]2 years ago

5 0

Answer:

$\fbox{3.81 \: grams}$

Explanation:

First consider the mol to mol ratio, the mol of a substance is simply the count of atoms in respect to avagadros number (approx. 6.02 × 10²³ molecules) in the period table. 1 mol of an element is simply it's mass count in the periodic table.

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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

If a substance conducts heat easily, it is considered to be a

GrogVix [38]

It is considered to be a conductor

5 0

3 years ago

Read 2 more answers

Consider the half reactions below for a chemical reaction.

ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0

2 years ago

In the presence of a base, blue litmus paper will . O turn purple stay blue to turn red

Inessa05 [86]

Answer:

In the presence of a base, blue litmus paper will turn red........

7 0

2 years ago

Read 2 more answers

Calculate the number of moles CuCl2 necessary to make 50 mL of a 0.15M solution

tekilochka [14]

Volume ⇒ 50 mL in liters : 50 / 1000 = 0.05 L

Molarity of solution ⇒ 0.15 M

Number of moles:

n = M * V

n = 0.15 * 0.05

n = 0.0075 moles of CuCl2

hope this helps!.

3 0

2 years ago