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kherson [118]
3 years ago
11

A sample of copper powder is heated in an evaporating dish using a Bunsen burner. If the mass of the powder in the evaporating d

ish increases after heating, this indicates that A. the copper must have combined with another substance. B. the copper must have lost matter. C. hot copper is more dense than cold copper. D. hot copper weighs more than cold copper.
Chemistry
1 answer:
Charra [1.4K]3 years ago
5 0

Answer : The correct option is, (A) the copper must have combined with another substance.

Explanation :

According to the question, when a sample of copper powder is heated in an evaporating dish by using Bunsen burner then the mass of the powder in the evaporating dish increases after heating that means the copper powder will be combined with the another substance.

If the mass of powder in the evaporating dish is decreases after the heating then the copper powder must have lost some matter.

The given option C and D are wrong statements because the hot copper is less dense than the cold copper and as we know that the density is directly proportional to the mass of a substance that means the hot copper weighs less than cold copper.

Hence, the correct option is, (A) the copper must have combined with another substance.

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A sample of C3H8 has 6.72 x 10^24 H atoms. <br> What is the total mass of the sample?
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The answer is 492.8 g


1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample. 

1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole    

6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n

n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol


2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:

Ar(C) = 12 g/mol

Ar(H) = 1 g/mol

Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol



3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:

m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g

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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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