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Tpy6a [65]
3 years ago
9

Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If

it is a mixture, classify it as homogeneous or heterogeneous.
a. Sweat
b. Carbon dioxide
c. Aluminum
d. Vegetable soup
Chemistry
2 answers:
ipn [44]3 years ago
7 0
A.Sweat-Mixture, homogeneous
b.Carbon Dioxide-Pure Substance, compound
c.Aluminium-Pure Substance, element
d.Vegetable Soup-Mixture, heterogeneous
koban [17]3 years ago
5 0

Explanation:

A pure substance is a substance that consists one or more atoms present in similar composition.

For example, aluminium, carbon dioxide, magnesium sulfate etc are all pure substances.

Whereas mixtures are the substances that contains two or more atoms mixed together in a different composition.

When molecules of solute are evenly distributed into the solvent then it is known as a homogeneous mixture.

For example, NaCl in water, sweat, sugar solution etc are all homogeneous mixtures.

When molecules of a solute are non-uniformly distributed then it is known as a heterogeneous mixture.

For example, sand in water, vegetable soup etc are heterogeneous mixtures.

Thus, we can classify the given substances as follows.

a) Sweat - homogeneous mixture

b) Carbon dioxide - pure substance

c) Aluminium - pure substance

d) vegetable soup - heterogeneous mixture

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lubasha [3.4K]

Answer:

26.2g = Mass of water in the calorimeter

Explanation:

The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:

Q = m*C*ΔT

<em>Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature</em>

<em />

Replacing we can write:

m_{metal}*C_{metal}*dT_{metal}=m_{water}*C_{water}*dT_{water}

13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)

240.9J = m(H₂O) * 9.2J/g

<h3>26.2g = Mass of water in the calorimeter</h3>
4 0
3 years ago
A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency
saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

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Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J =  6.63×10⁻³⁴ Js ×f

f =  5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0
3 years ago
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Verdich [7]

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Explanation:

Mark me as a brainliest happy po ako sa pag sagot sana tama ako kung mali paki. Comment lang dito kasi solve ako ulit

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3 years ago
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Answer:

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In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.

Explanation:

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