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yan [13]
3 years ago
13

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.

Chemistry
1 answer:
MrRissso [65]3 years ago
6 0
<h2>Answers to the rest of the assignment:</h2>

**Check for proof photos at the bottom.**

__________________________________________________________

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.  

Cranberry juice at the store

A. homogeneous

C. solution

Smoke  

B. heterogeneous  

D. colloid

__________________________________________________________

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.  

Blood

B. heterogeneous  

E. suspension

Salad dressing

B. heterogeneous  

E. suspension

__________________________________________________________

Shown here is a person shaving.  Under magnification, the shaving foam might look like the image above the shaver. What type of mixture does the foam demonstrate? Give your reasoning.

Foam is a colloid. Colloids includes gas dispersed in a liquid and it also includes gas dispersed in a solid

On the next slide, you can select any answers you want, or you can select nothing. There's no wrong answer.  

__________________________________________________________

A student squeezes several oranges to make a glass of orange juice. The juice contains pieces of orange pulp mixed with the juice. Explain why this drink can be considered a combination of a suspension and a solution.

The juice contains sugars, plant pigments, and other chemicals dissolved in water. This is a solution. The pieces of orange pulp will rise to the top or settle to the bottom of the juice if it is allowed to sit. The pieces of pulp mixed with the juice form a suspension

On the right side, you can also select any answers or none at all. There is no wrong answer.

__________________________________________________________

<h2>Explanation:</h2>

There are two main types of mixtures, homogeneous mixtures and heterogeneous mixtures. The components of a homogeneous mixture are evenly distributed throughout the mixture. The properties of a mixture are the same everywhere. The components of a heterogeneous mixture are not evenly distributed. Different regions of this mixture have different   properties.

Particles in a homogeneous mixture do not settle down or separate when left alone. Particles in a heterogeneous mixture eventually separate or settle when left alone.

  • Colloids are a type of heterogeneous mixture that do not naturally settle out quickly, and can be separated with different methods.
  • A suspension is another type of heterogeneous mixtures in which the components of the mixture will quickly settle out or can be filtered or separated.

Here are photos of Edge just incase.

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5 0
3 years ago
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Write electron configurations for Gallium, Ga (Z=31), and show the total valence electrons
ivann1987 [24]

<u>Answer:</u> The electronic configuration of gallium is written below and number of valence electrons is 3.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

Valence electrons are defined as the electrons present in the outermost shell of an atom.

We are given:

An element Gallium having atomic number as 31.

Number of electrons = 31

Electronic configuration of Gallium is: (Z=31):1s^22s^22p^63s^23p^64s^23d^{10}4p^1

This element has 3 electrons in its outermost shell. So, the number of valence electrons is 3

Hence, the electronic configuration of gallium is written below and number of valence electrons is 3.

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4 years ago
They say that the sun will blow up after 200 billion years.
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An element is made up from ?
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370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is
NikAS [45]

Answer : The final temperature of the mixture is 61.4^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = c_2 = specific heat of water = same

m_1 = m_2 = mass of water  =  same

\rho_1 = \rho_2 = density of water = 1.0 g/mL

V_1 = volume of water at 80.0^oC  = 370cm^3=370mL

V_2 = volume of water at 4^oC  = 120cm^3=120mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of water = 80.0^oC

T_2 = initial temperature of water = 4^oC

Now put all the given values in the above formula, we get:

(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)

(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC

T_f=61.4^oC

Therefore, the final temperature of the mixture is 61.4^oC

6 0
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