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3 years ago

Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g

1 answer:

4 0

Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

$gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl$

Have a nice day!

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What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

RideAnS [48]

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.0 \times {10^{15}}\;{\text{Hz}}$ is $\boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}$

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

${\mathbf{E=h\nu }}$ ......(1)

Here,

${\text{E}}$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The frequency of UV rays is $1.0 \times {10^{15}}\;{\text{Hz}}$ or $1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}$

The value of Planck’s constant is $6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}}$ .

Substitute these values in equation (1)

$\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}$

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is ${\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}}$ .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

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svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

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1 mol

temperature = 100°C

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$(P + \frac{a}{v^{2}} )(v - b) = RT$

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P = 1.336 - 0.0094

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