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olga55 [171]
3 years ago
9

Which of the following is true about homogeneous mixtures? A. They are known as solutions

Chemistry
2 answers:
Natali [406]3 years ago
8 0

Answer: Option (A) is the correct answer.

Explanation:

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A homogeneous mixture is also known as solution as a homogeneous mixture is basically a clear solution.

Composition of the solute and solvent particles can vary with time or set of conditions.

A homogeneous mixture can be present any state.

Thus, we can conclude that the statement they are known as solutions is true about homogeneous mixtures.

Helga [31]3 years ago
7 0

Answer:

A is your answer..... !!!!!!!!!!!!!

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Thomas Melvill

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A typical air sample in the lungs contains oxygen at 100 mmhg, nitrogen at 573 mmhg, carbon dioxide at 40 mmhg, and water vapor
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What is the function of this instrument?
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To measure weight of a item
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4 years ago
For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea
BlackZzzverrR [31]

Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

6 0
3 years ago
certain gas occupies 6.24L at pressure of 760 mm Hg.If the pressure is reduced to 60.0mm Hg,what would the new volume be?​
vovikov84 [41]

Answer:

79.04 L

Explanation:

We are given;

Initial Volume; V1 = 6.24L

Initial Pressure; P1 = 760 mm Hg

Final pressure; P2 = 60.0mm Hg

To solve for final volume, we will use Boyles law;

P1•V1 = P2•V2

Let's make V2 which is the final volume the subject;

V2 = (P1•V1)/P2

V2 = (760 × 6.24)/60

V2 = 79.04 L

3 0
3 years ago
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