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3 years ago

A well-mixed mixture containing a solvent and dissolved a solute.

Chemistry

1 answer:

baherus [9]3 years ago

8 0

Answer:

Solution

Explanation:

A solution is a well-mixed mixture containing a solvent and dissolved solute.

It is termed a homogenous mixture.

Solutions are homogeneous mixtures of solutes and solvents. In a solution, the solute particles are distributed uniformly in the solvent. The solute is the substance that is dissolved to make the solution

It is usually present in smaller amount than the solvent. The solvent is the dispersing substance that allows the solute to go into solution.

The solute could be a solid or liquid or gas but usually a liquid. A solution in which the solvent is water is an aqueous solution.

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SEP Analyze Data Use graphing software or draw a graph from the data in the table. Label "Heat input (joules)" on the y-axis and

galben [10]

The characteristics of the graphic representation are to find the scales and make the best graphic, in the attachment we have the points plotted

The graphical representation is one of the best methods to see the relationships between groups of data and to be able to find the functional relationships between them.

In this case, it is requested to make a scatter plot where the measured temperature is placed on the x-axis and the heat on the y-axis.

In the attachment we can see a graph with the requested data, the most important part of finding these graphs is looking for the scales

x-axis

scale x = $\frac{\Delta Values}{paper length}$

y-axis

scale y = $\frac{\Delta Values}{paper \ lengthy}$

In general, the paper is 20 x 30 cm, we select the shortest part as the x axis

scale x = $\frac{50-30}{20}$

scale x = 1º $\frac{ C}{cm \ peper}$

Let's find the scale on the axis and the maximum value is 54000 and the minimum value is 3800, as the minimum value is much less than the maximum value, let's set this value to zero

scale y = $\frac{54000-0}{30}$

scale y = 1800 $\frac{j}{cm \ paper}$

To facilitate marking the values, we select a slightly larger scale,

Selected scale

scale y = 2000 \frac{j}{cm \ paper}

Let's mark the points.

From observing these graph we see;

The relationships are linear
For the same material the amount of heat necessary to reach the same temperature is proportional to the mass of the material

Using the characteristics of the graphical representation, it is possible to find the scales and make the best graph

Learn more here: brainly.com/question/12511806

6 0

3 years ago

Heat energy that is lost cannot usually be used to produce work.<br><br> True False

Veseljchak [2.6K]

Answer:

true

Explanation:

thanks for free point

7 0

3 years ago

What are the 6 steps to solving a mass-mass problem

aev [14]

Answer:

Write a balanced equation for the reaction.

Write the given mass on a factor-label form.

Convert mass of reactant to moles of reactant.

Convert moles of reactant to moles of product.

Convert moles of product to grams of product.

Pick up the calculator and do the math.

Explanation:

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4 0

4 years ago

Suppose 'A' is a liquid aromatic compound with molecular weight 78 and burns with sooty flame. a.Give the name of the compound '

Oliga [24]

Aromatic compounds are compounds that contain carbon-carbon multiple bonds.

The question did not mention that a heteroatom is present in the compound so we can assume that there is none of such. In that case, the compound contains only hydrogen and carbon.

So,

(CH)n = 78

where n is the number of each atom present.

(12 +1)n = 78

n = 78/13

n = 6

The molecular formula of the compound is C6H6

When C6H6 is treated with .conc.HNO3/conc.H2SO4 the compound shown in image 1 is formed. The reaction occurs at the C-C multiple bond.

When C6H6 is reacted with chlorine in the presence of sunlight, hexachlorobenzene (shown in image 2 attached) is formed.

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3 0

3 years ago

The reaction rate is k[Ce4+][Mn2+] for the following reaction: 2Ce4+(aq) + Tl+(aq) + Mn2+(aq) → 2Ce3+(aq) + Tl3+(aq) + Mn2+(aq W

sdas [7]

Answer:

Manganese (II) ion, Mn²⁺

Explanation:

Hello,

In this case, given the overall reaction:

$2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)$

Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺

Best regards.

8 0

4 years ago