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liberstina [14]

4 years ago

15

The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 9.00×1010 atoms of zinc e

mitting light in the instrument flame per second, what energy (in joules) must the flame supply during this time to achieve this level of emission?

1 answer:

KengaRu [80]4 years ago

3 0

Answer:

$8.4\cdot 10^{-8}J$

Explanation:

The energy emitted by a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

For the photons emitted by the zinc atoms,

$\lambda = 214 nm = 214 \cdot 10^{-9} m$

So the energy of a single photon emitted is

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{214\cdot 10^{-9}}=9.3\cdot 10^{-19}J$

And since the number of atoms is

$N=9.0\cdot 10^{10}$

The total energy emitted will be

$E=NE_1 = (9.0\cdot 10^{10})(9.3\cdot 10^{-19})=8.4\cdot 10^{-8}J$

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A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg/s of cool water at 25 °C with 0.8 kg/s of

Tamiku [17]

Explanation:

The given data is as follows.

$m_{1}$ = 0.8 kg/s = 800 g, $C_{1} = 4.18 J/g^{o}C$

$m_{2}$ = 1 kg = 1000 g, $T_{1}$ = (75 - T),

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Now, according to the law of conservation of energy,

$m_{1}C_{1}T_{1} = m_{2}C_{2}T_{2} + mL$

$800 \times 4.18 \times (75 - T) = 1000 \times 4.18 \times (T - 25)$ + 30000

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3 years ago

All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outle

VARVARA [1.3K]

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage ......................1

here voltage is same in all three device

so

current by toaster is

I = $\frac{1140}{120}$

I = 9.5 A

and

current by blender

I = $\frac{270}{120}$

I = 2.25 A

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current by lamp is

I = $\frac{80}{120}$

I = 0.667 A

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The value of the force, F₀, at equilibrium is equal to the horizontal

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Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

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Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

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