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mixer [17]
2 years ago
14

Give the atomic symbol for an element that does not exist as a molecule or extended structure. that is, an element that exists o

nly as discrete atoms.
Chemistry
2 answers:
Bezzdna [24]2 years ago
6 0

Generally an elements exist as a molecule of extended structure as it can form bond with the atom / atoms of same element

They form bond in order to full fill their octet / duet (o=in case of hydrogen) and to attain stability

It is considered that the molecules with complete octet of atom is stable

However, in case of noble gases (like helium, neon , argon, Krypton etc) they already have that complete octet and they have already attained that stability.

So they do not undergo bond formation with the atoms of same type hence they do not form He2 like H2

Or we can say that these elements  ( He , Ne and Ar) exists as discrete atom

RUDIKE [14]2 years ago
4 0
An element that exist as a discreet atom has only one atom which can stand alone on its own. Example of this is Neon, which is a noble gas. The chemical symbol for neon is Ne.
Elements are able to exist as discreet atom because they are chemical stable and inert.
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Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F
nirvana33 [79]

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.  If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

  • <u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is False.

  • <u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)

Ionic form of the above equation follows:

Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

  • <u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is True.

  • <u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the correct answer is False.

  • <u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the correct answer is True.

3 0
2 years ago
The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

The density of O₂ gas is 1.71 \frac{g}{L}

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

  • P=1.27 atm
  • molar mass of O₂= 32 \frac{g}{mol}.
  • R= 0.0821 \frac{atm*L}{mol*K}
  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
2 years ago
The requirements for one type of atom to substitute for another in a solid solution are:
Zinaida [17]

Answer: Pine trees, strawberry plants, and cacti are all plants. Name and describe three or more characteristics that you think scientists could use to separate these members of the Plantae kingdom into smaller groups.

Answer:

Explanation: Pine trees, strawberry plants, and cacti are all plants. Name and describe three or more characteristics that you think scientists could use to separate these members of the Plantae kingdom into smaller groups.

Answer:

Pine trees, straw Pine trees, strawberry plants, and cacti are all plants. Name and describe three or more characteristics that you think scientists could use to separate these members of the Plantae kingdom into smaller groups.

Answer:

erry plants, and cacti are all plants. Name and describe three or more characteristics that you think scientists could use to separate the Pine trees, strawberry plants, and cacti are all plants. Name and describe three or more characteristics that you think scientists could use to separate these members of the Plantae kingdom into smaller groups.

Answer:

se members of the Plantae kingdom into smaller groups.

Answer:

Pine trees, strawberry plants, and cacti are all plants. Name and describe three or more characteristics that you think scientists could use to separate these members of the Plantae kingdom into smaller groups.

Answer:

5 0
2 years ago
Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio
AlexFokin [52]

Answer:

The equilibrium concentration of CH₃OH is 0.28 M

Explanation:

For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)

The equilibrium constant (Keq) is given for the following expresion:

Keq= \frac{(CH3OH)}{(CO) x (H2)^{2}} =14.5

Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.

We have:

(CH3OH)= ?

(CO)= 0.15 M

(H2)= 0.36 M

So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:

14.5= \frac{(CH_{3}OH) }{(0.15 M) x (0.36 M) ^{2} }

14.5 x (0.15 M) x (0.36)^{2} = (CH₃OH)

0.2818 M = (CH₃OH)

6 0
2 years ago
What is the molarity of a solution that is made by adding 57.3 g of MgO to 500.0 mL of solution?
kifflom [539]

Answer:

2.843 M

Explanation:

Molarity = moles / volume

<u>Milliliters to liters:</u>

500 mL <u>= .500 L</u>

<u>Grams to moles:</u>

MgO molar mass = 16.00 + 24.31 = 40.31 g/mol

57.3 g x 1 mol / 40.31 g = <u>1.421 mol</u>

<u>Molarity:</u>

1.421 mol / .500 L =  2.843 M

0 0
2 years ago
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