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3 years ago

Which set is an example of an element, a compound, and a mixture?

Chemistry

1 answer:

Nitella [24]3 years ago

4 0

Answer:

Element: Sulfur ; S

Compound: Water ; H2O

Mixture: Gunpowder ; 2 KNO3 + S + 3 C -> K2S + N2 + 3 CO2

Explanation:

An element is a pure substance that cannot be broken down into simpler substances. A periodic table for example has elements.

A compound is a chemical substance that's made of multiple elements held together by chemical bonds. All elements from the periodic table can be combined to make a compound but they have to bond a specific way in order for them to actually make something. Water is the easiest to remember.

A mixture is two or more substances that are physically combined. They are not chemically combined though. There are a few types of mixtures including homogeneous, heterogeneous, etc. Gunpowder is actually a mixture. It contains sulfur, carbon and potassium nitrate.

Don't be afraid to reach out with further questions, I hope this helps!

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Use the link Standard Reduction Potentials. Write net equations for the spontaneous redox reactions that occur during the follow

GalinKa [24]

Answer:

a) Fe(s) + Ni^2+(aq) ----> Fe^2+(aq) + Ni(s)

b) no reaction

c) no reaction

d) 2Mg(s) + 2H2O(l)-----> 2Mg^2+(aq) + O2(g) +4H^+(aq)

e) no reaction

Explanation:

It is important to say here that the ability of a particular chemical specie to displace another chemical specie is dependent on the relative standard reduction potentials of the species involved.

All the reactions stated above are redox reactions. Let us take reaction E as an example. Mg^2+ has a reduction potential of -2.37 V while Cr^3+ has a reduction potential of -0.74V. Since the reduction potential of magnesium is more negative than that of chromium, there is no reaction when a piece of chromium metal is dipped into a solution of Mg^2+.

Similarly, though metals displace hydrogen gas from dilute acids, metals that are less than hydrogen in the reactivity series cannot do that. This explains why there is no reaction when copper and silver are dipped into dilute acid solutions.

Reaction occurs when iron is dipped into a nickel solution because the reduction potential of Fe^2+ is far more negative than that of Ni^2+.

7 0

3 years ago

An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density

V125BC [204]

Answer:

Explanation:

For the density of a face-centered cubic:

$Density = \dfrac{4 \times M_w}{N_A \times a^3}$

where

$M_w$ = molar mass of the compound

$N_A=$ avogadro's constant

$a^3 =$ the volume of a unit cell

Given that:

Density $(\rho)$ = 19.30 g/cm³

a = 0.408 nm

a = $0.408 \times 10^{-9} \times 10^{2} \ cm$

a = $4.08 \times 10^ {-8} \ cm$

∴

$19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}$

$M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}$

$M_w=197.37 \ g/mol$

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0

3 years ago

Which of the following redox reactions do you expect to occur spontaneously in the forward direction?

Leni [432]

Among the choices provide above the redox reactions do you expect to occur spontaneously in the forward direction is the below:

Fe2+(aq) + Zn(s) -> Fe(s) + Zn2+(aq)
 Al(s) + 3Ag+(aq) -> Al3+(aq) + 3Ag(s)

Those reactions will proceed when the metal that is a solid is higher up in the electromotive series than the metal that is an ion (dissolved).

5 0

3 years ago

Read 2 more answers

Give the electron confiuration of Chromium (Cr), atomic number 24.

nadezda [96]

Chromium is a transition metal and it has 24 electrons and here is the orbital diagram. If we're going to make this short hand and make the electron configurationfor this we would make this 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d4 okay from now on every time you see 3d4 you're going to change it, we do not like 3d4.

4 0

4 years ago

Read 2 more answers

Federal regulations set an upper limit of 50 parts per million (ppm) of NH3 in the air in a work environment [that is, 50 molecu

pogonyaev

Answer:

1) A total of 5.91×10-3 grams were drawn into the HCl solution

2) 84.3 ppm of NH3 were in the air

3) As 84.3 ppm is higher than 50 ppm established in the regulation, the manufacturer does not comply with it.

Explanation:

The problem shows the following process: an amount of air with NH3 passes through a HCl solution. The NH3 reacts with HCl reducing the concentration of the latted and finally the remaining HCl is titrated with NaOH.

The process to solve this problem should go as follows:

a) Calculate the amount of the remaining HCl that was titrated with the NaOH at the end.

b) Calculate the amount of HCl that reacted with NH3, using the data from a)

c) Calculate the amount of NH3 present in air using the data from b)

d) Calculate the grams of NH3 using the data from c) to solve question 1)

e) Calculate the number of moles of air

f) Calculate the ppm of NH3 in air using the data from c) and e) to solve questions 2) and 3)

So, let's proceed:

a) To do this we need to take a look at the chemical equation of the HCl and NaOH reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

And we see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, being n the number of moles:

n(NaOH) = n(HCl)

(5.86×10-2 M)×(14.5×10-3 L) = n(HCl)

n(HCl) = 8.50×10-4 moles of HCl

b) So, as we now have the amount of the remaining HCl, we need to find out how much HCl reacted with NH3 in the first place, so we need to substract the number of moles found in a) from the number of moles we had initially:

n(HCl)_reacted with NH3 = n(HCl)_initial - n(HCl)

n(HCl)_reacted with NH3 = (1.13×10-2 M)×(106×10-3 L) - 8.50×10-4

n(HCl)_reacted with NH3 = 3.49×10-4 moles of HCl

c) Now, as we know the amount of HCl that reacted with NH3, we can calculate the amount of NH3 that was drawn into the solution using the chemical equation (fortunately, the equation is already balanced):

NH3(aq) + HCl(aq) → NH4Cl(aq)

And we can see 1 mole of NH3 reacts with 1 mole of HCl, so we can conclude that 3.49×10-4 moles of HCl have reacted with 3.49×10-4 moles of NH3.

As we are being asked by the grams, we must convert that using the molar mass of NH3 that is 17 g/mol (N=14, H=1), so:

grams of NH3 = (3.49×10-4 mol)×(17 g/mol) = 5.91×10-3 grams of NH3

d) Now we must calculate the number of moles of air in order to be able to calculate the parts per million of NH3:

In this case we have to notice that we have passes air at a rate of 10.0 liters per minute and we have done it by 10 minutes, that means that the total amount of air (in liters) we have passed through the solution is:

liters of air = 10 min × 10 L/min = 100 L

e) That volume of air can be converted into moles using the information from question 2):

moles of air = (100 L) × (1.2 g/L) × (1 mol/29 g) = 4.14 moles

f) We can calculate now using the information from c) and e) as follows:

ppm of NH3 in air = number of moles of NH3 / number of moles of air × 1000000

ppm of NH3 = (3.49×10-4 mol)/(4.14 mol)×1000000 = 84.3 ppm

In conclusion, the manufacturer does not comply with the regulation of maximum 50ppm of NH3.

6 0

4 years ago