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3 years ago

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On the basis of the expected charges on the monatomic ions, give the chemical formula of each of the following compounds: (a) ma

gnesium arsenide; (b) indium(III) sulf de; (c) aluminum hydride; (d) hydrogen telluride; (e) bismuth(III) f uoride

1 answer:

lakkis [162]3 years ago

5 0

Answer:

(a) magnesium arsenide: As₂Mg₃

(b) indium(III) sulfide: In₂S₃

(c) aluminum hydride: AlH₃

(d) hydrogen telluride: H₂Te

(e) bismuth(III) fluoride: BiF₃

Explanation:

(a) magnesium arsenide:

The expected oxidation states are-

Magnesium ion- Mg²⁺

Arsenide ion- As³⁻

The chemical formula is: As₂Mg₃

(b) indium(III) sulfide:

The expected oxidation states are-

Indium ion- In³⁺

Sulfide ion- S²⁻

The chemical formula is: In₂S₃

(c) aluminum hydride:

The expected oxidation states are-

Aluminium ion- Al³⁺

Hydride ion- H⁻

The chemical formula is: AlH₃

(d) hydrogen telluride:

The expected oxidation states are-

Hydrogen - H⁺

telluride ion- Te²⁻

The chemical formula is: H₂Te

(e) bismuth(III) fluoride:

The expected oxidation states are-

bismuth ion- Bi³⁺

fluoride ion- F⁻

The chemical formula is: BiF₃

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The lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group (red). in this rep

Fed [463]

The question is incomplete, the complete question is;

The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group______ (red).

In this representation, each Y atom needs ______ electron(s) to complete its octet, and gains these electrons by forming______ bond(s) with atoms of H .

There are ______ unshared electron pair(s) and _______bonding electron pair(s) in the product molecule.

The bonds in the product are _________ (Ionic or Covalent)

Answer:

1) 16

2) 2 electrons

3) 2 bonds

4) 2 unshared pairs of electrons

5) 2 bonding pairs of electrons

6) The bonds in the product are covalent

Explanation:

Group sixteen elements have six electrons on their outermost shell. These include two unshared pairs of electrons and two unpaired electrons. These two unpaired electrons can now be covalently bonded to two hydrogen atoms to give H2Y. The compound H2Y has two lone pairs and two bond pairs of electrons.

H2Y can be a general formula for all hydrides of group 16. They are all very similar in structure but gradually differ in physical and chemical properties according to the graduated variation observed down the group.

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3 years ago

Write the name of the compounds

blagie [28]

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CuO = Copper (II) oxide

NH4CI = Ammonium Chloride

Mn(OH)2 = Pyrochroite

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CIO2 = Chlorine dioxide

NaF = Sodium fluoride

FeSO3 = Iron (II) Sulfite

Fe(NO3)3 = Iron (III) Nitrate

Cr(NO2)3 = Chromium (III) Nitrite

NaHCO3 = Sodium Hydrogen Carbonate

H2PO4 = Dihydrogen Phosphate Ion

NaCN = Sodium Cyanide

IF7 = Iodine Heptafluoride

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5 0

3 years ago

How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250

Grace [21]

<u>Answer:</u> The number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.034mol/L.atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

$C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M$

To calculate the number of moles for given molarity, we use the equation:

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

Molarity of carbon dioxide = $8.5\times 10^{-5}M$

Volume of solution = 0.550 L

Putting values in above equation, we get:

$8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, $4.675\times 10^{-3}$ moles of carbon dioxide will contain = $(6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}$ number of molecules

Hence, the number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

3 0

3 years ago

Read 2 more answers

Calculate Ecell at 80 ºC for a voltaic cell based on the following redox reaction: H2(g, 1.25 atm) + 2AgCl(s) → 2Ag(s) + 2H+(aq,

iris [78.8K]

Answer:

Ecell = +0.25V

Explanation:

the half-cell reactions for a voltanic cell

cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)

anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻

we have the standard cell potential E⁺cell = 0.18V at 80C respectively

Q = [H⁺]/[Cl⁻]

sub for [H+] = 0.10M and [Cl-] = 1.5M

Q= 0.1M/1.5M

Q = 0.067

Ecell = E⁺cell - $\frac{0.059}{n}$ logQ

= 0.18 - $\frac{0.056}{1}$ log 0.067

0.18- 0.059(-1.174)

Ecell = +0.25V

6 0

3 years ago

En un experimento hacemos reaccionar 12 g de carbono con 32 g de oxígeno para formar dióxido de carbono. Razona si podemos saber

max2010maxim [7]

Answer:

La masa de óxido de carbono iv formado es 44 g.

Explanation:

En esta pregunta, se nos pide calcular la masa de óxido de carbono iv formado a partir de la reacción de masas dadas de carbono y oxígeno.

En primer lugar, necesitamos escribir una ecuación química equilibrada.

C + O2 → CO2

De la ecuación, 1 mol de carbono reaccionó con 1 mol de oxígeno para dar 1 mol de óxido de carbono iv.

Ahora, si marca las masas en la pregunta, verá que corresponde a la masa atómica y la masa molar de la molécula de carbono y oxígeno, respectivamente. ¿Qué indica esto?

Como tenemos una relación molar de 1: 1 en todo momento, lo que esto significa es que la masa de óxido de carbono iv producida también es la misma que la masa molar de óxido de carbono iv.

Por lo tanto, procedemos a calcular la masa molar de óxido de carbono iv Esto es igual a 12 + 2 (16) = 12 + 32 = 44 g Por lo tanto, la masa de óxido de carbono iv formado es 44 g

5 0

3 years ago