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Minchanka [31]
3 years ago
5

Which one of the following is not a homogeneous mixture?. A. vegetable soup. B. carbonated beverages. C. saline solution. D. bre

wed tea
Chemistry
2 answers:
VashaNatasha [74]3 years ago
8 0
Vegetable soup is the one among the following options given in the question that is not a homogeneous mixture. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer you were looking for and it has helped you.
IgorC [24]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

When there are two or more substances mixed together in such a way that their particles are evenly distributed throughout the mixture then it is known as a homogeneous mixture.

For example, saline solution, carbonated solution, brewed tea etc are all homogeneous solutions.

But when the solute particles are not evenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, vegetable soup is a heterogeneous mixture.

thus, we can conclude that out of the given options vegetable soup  is not a homogeneous mixture.

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When liquid water changes into solid ice, it increases in ?
kramer

when liquid water changes into solid ice, it increases in mass

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3 years ago
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suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20 C. What would have been the fi
balandron [24]

Answer:

The final temperature of water is 54.5 °C.

Explanation:

Given data:

Energy transferred = 65 Kj

Mass of water = 450 g

Initial temperature = T1 = 20 °C

Final temperature= T2 = ?

Solution:

First of all we will convert the heat in Kj to joule.

1 Kj = 1000 j

65× 1000 = 65000 j

specific heat of water is 4.186 J /g. °C

Formula:

q = m × c × ΔT

ΔT = T2 - T1

Now we will put the values in Formula.

65000 j = 450 g × 4.186 J /g. °C  × (T2 - 20°C )

65000 j = 1883.7 j /°C × (T2 - 20°C )

65000 j/ 1883.7 j /°C  = T2 - 20°C

34.51 °C = T2 - 20°C

34.51 °C + 20 °C = T2

T2 = 54.5 °C

5 0
4 years ago
A scientist introduces an unknown type of particle into a cathode ray tube. The unknown particles travel in the same direction a
sergij07 [2.7K]

Answer: B

Explanation:

5 0
3 years ago
An experiment is carried out to observe the melting ice. Heat energy is supplied at a constant rate. The temperature of the ice
zhenek [66]

Answer:

If the ambient temperature around a piece of ice increases, the temperature of the ice will increase as well. However, this steady increase in temperature stops as soon as the ice reaches its melting point. At this point, the ice undergoes a change of state and turns into liquid water, and its temperature won't change until all of it has melted. You can test this with a simple experiment. Leave a cup of ice cubes in a hot car and monitor the temperature with a thermometer. You'll find that the icy water remains at a frosty 32 degrees Fahrenheit (0 degrees Celsius) until all of it has melted. When that happens, you'll notice a quick temperature rise as the water continues to absorb heat from the inside of the car.

7 0
3 years ago
What is the molar mass of 4.23 g of an elemental gas in a 2.5L container at 282K and 1.4 atm?
balu736 [363]

Answer:

27.98g/mol

Explanation:

Using ideal gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

T = temperature (K)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

According to the information given:

V = 2.5L

P = 1.4 atm

T = 282K

n = ?

Using PV = nRT

n = PV/RT

n = 1.4 × 2.5/0.0821 × 282

n = 3.5/23.1522

n = 0.151mol

Using the formula to calculate molar mass of the elemental gas:

mole = mass/molar mass

Molar mass = mass/mole

Molar mass = 4.23g ÷ 0.151mol

Molar mass = 27.98g/mol

6 0
3 years ago
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