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4 years ago

How is matter classified as substances elements compounds and mixtures?

1 answer:

4 0

Matter is classified as substances and mixtures.
Substances include elements and compounds. Element is something that cannot be broken down into anything more simple even by chemical methods. There are over 100 elements on earth and the most common ones include oxygen or aluminum or silicon.
Compound is something where 2 or more elements are chemically combined together, such as by electrolysis or heating. It cannot be broken down by simple physical methods, such as evaporation or filtration. Common Examples include carbon dioxide, sodium chloride etc.
Mixture is where elements or compounds are mixed together, and can be separated by simple physical methods. Examples include air or sea water. The proportion of the elements and compounds are not fixed.

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The spontaneous reaction that occurs when the cell in the picture operates is as follows: 2Ag+ + Cd(s) ???? 2 Ag(s) + Cd2+ (A) V

Snowcat [4.5K]

Answer:

14. B 15. D 16. C 17. B

Explanation:

The spontaneous reaction that occurs when the cell operates is shown below:

$2Ag^{+} + Cd_{(s)}$ ⇒ $2Ag_{(s)} + Cd^{2+}$

We need to select the correct option from the list below for the following questions.

(A) Voltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.

14. A 50-milliliter sample of a 2-molar $Cd(NO_{3})_{2}$ solution is added to the left beaker.

If a 50-milliliter sample of a 2-molar $Cd(NO_{3})_{2}$ solution is added to the left beaker, the voltage decreases but its value remains greater than zero. The correct option is B

15. The silver electrode is made larger.

If the silver electrode is made larger, no change in the value of the voltage since we don't have the idea of the initial value. The correct option is D.

16. The salt bridge is replaced by a platinum wire.

If the salt bridge is replaced by a platinum wire, there will be no passage of electrons because electrons can't pass through a platinum wire. Therefore, the voltage will be zero and remains at zero. The correct option is C.

17. Current is allowed to flow for 5 minutes.

If current is allowed to flow for 5 minutes, the voltage decreases but its value remains greater than zero. The correct option is B.

7 0

4 years ago

A temperature increase causes the particles to what ??!

lord [1]

To move around more. bounce off walls ect. they are more energised

6 0

3 years ago

Connect and Apply scientific knowledge to explain the following situations:

mixer [17]

Answer:

b.Rearrange the two equations to make each variable the focus (you will end up with three variations for each equation).

7 0

2 years ago

When cations and anions join, they form what

uranmaximum [27]

Answer:

ionic

Explanation:

6 0

3 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

3 years ago