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2 years ago

What is a diffusion, suspension, and colloid?

1 answer:

4 0

Answer: Colloidal suspensions are dispersions of small particles, ranging in size from 1 nanometer to 1 micrometer, in a solvent. ... This action prevents the colloid particles from settling out of the solvent. More importantly, Brownian motion contributes to the diffusion of colloid particles through the solution.

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What is produced during the replacement reaction of Cu(NO3)2 and Zn?

Sidana [21]

Zn(NO3)2 + Cu will be the result of this single displacement (replacement) reaction :))

5 0

3 years ago

Read 2 more answers

When harmful bacteria invade a wound, the skin around the wound turns red. This is known as:

nasty-shy [4]

Answer:

Inflammation

Explanation:

Inflammation is the response to a cut. The blood vessels expand when you get cut to allow white blood cells to flock to the wound site. These attack any bacteria that got past the clot. If you mean <u>excess</u> bacteria, then that would be an infection.

4 0

3 years ago

Which of these statements describes a chemical property

sp2606 [1]

Sugar burns in air describes a chemical property

<h3>Further explanation </h3>

In the matter of a substance, there is a process of change.

2 changes occur in the mater, namely: physical changes and chemical changes

Classification uses the principle of the initial and final state of the substance.

Physical changes are changes in the matter which in the process do not cause new substances. So the properties of the particles remain the same.

Generally, the physical changes that occur are changes in shape / size and changes due to heating, or changes in volume

Example:

melting

a solid changes to a liquid

freezing

a liquid changes to a solid

boiling

a solid changes to a liquid

Chemical changes are changes in the process that produce new substances that are different from the initial / forming substances. Chemical changes are usually known as chemical reactions. So there are products / reactions and reactants

So in chemical changes, there is a change in the molecule / particle

From this we can know the physical and chemical properties of a substance

physical properties:

melting point, viscosity, ductile, freezing point, boiling point, density, odor, solubilty, turbidity, color etc.

chemical properties:

Highly flammable, corroded, explosive, toxic etc.

Let's look at the answer choices in the attached picture

1. water boils at 100 C

physical properties of water that is boiling point

2. mercury is liquid at room temperature

physical properties: the form of substances

3. Helium is the lightest monoatomic element

physical properties: the weight or mass of a substance

4. sugar burns in water to form water and carbon dioxide

chemical properties: the formation of new substances

<h3>Learn more </h3>

an example of a physical change

brainly.com/question/1119909

example of a chemical change brainly.com/question/9388643

examples of chemical changes brainly.com/question/11189422

7 0

3 years ago

Read 2 more answers

Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)

GarryVolchara [31]

$\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \; \text{kJ} \cdot \text{mol}^{-1}$

Explanation

$\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)$

$\Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)
$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)

As an the most stable allotrope under standard conditions, $\Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}$

By definition,

$\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products}) - \Delta H\textdegree{}_f (\text{all reactants})$

$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \; \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\ = \Delta H\textdegree{}_{\text{rxn}}$

$\begin{array}{ccc} \Delta H\textdegree{}_f (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}$

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the $\Delta H\textdegree{} _f$ given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

$\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \; \text{kJ} \cdot \text{mol}^{-1}$ (Cox, Wagman, et al., 1984)

... yields $\Delta H\textdegree{}_f (\text{IF} \; (g) ) \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}$ , which deviates significantly from the experimental value of $-94.76 \; \text{kJ}\cdot \text{mol}^{-1}$ (Chase, 1998.) It is thus assumed that the $\Delta H\textdegree{}_\text{rxn}$ value provided requires a reaction with $\text{I}_2 \; (s)$ rather than $\text{I}_2 \; (g)$ as a reactant.