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3 years ago

PLEASE HELP ME! WILL GIVE BRAINLEST! I need this ASAP!

Chemistry

1 answer:

Fofino [41]3 years ago

7 0

1. Al(C₂H₅)₂OCl₃

Al(C₂H₅)₂OCl₃ is made from 5 different elements. They are Al, C, H, O and Cl. Number of atoms in each element can be found as follows.

number of Al atoms = 1
number of C atoms = 2 x 2 = 4
number of H atoms = 5 x 2 = 10
number of O atoms = 1
number of Cl atoms = 3

Total number of atoms = 1 + 4 + 10 + 1 + 3 = 19

2. BCl₃

BCl₃ is a simple neutral molecule. It consists of 2 different elements as B and Cl. Total number of atoms is 4. Number of atoms can be calculated as follows.

number of B atoms = 1
number of Cl atoms = 3

Total number of atoms = 1 + 3 = 4

3. H₂O + H⁺

H₂O makes H₃O⁺ (hydronium ion) with H⁺ which is a positive ion. H₃O⁺ consists of two different elements as H and O. Number of atoms are as follows.

number of H atoms = 3
number of O atoms = 1

Total number of atoms = 3 + 1 = 4

You might be interested in

What would the molecular formula be if lithium and sulfur reacted to form a neutral compound?

skad [1K]

Answer:

D, Li2S

Explanation:

This is because Lithium, which is in group IA of the periodic table, has a charge of +1. Sulfur will have a charge of -2 because it is in group 6A in the periodic table, which means to balance these out, there needs to be 2 lithium ions which would result in a charge of +2. With Lithium now having a charge of +2 due to having two atoms in the compound, and sulfur already having a charge of -2 as one atom, these two cancel out meaning the compound is neutral.

4 0

2 years ago

2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)

defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = $\frac{number of moles}{volume}$

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

207.2 g of Pb ----------------- 1 mol

x g ----------------- 0.0013 moles

x = (0.0013 x 207.2) / 1

x = 0.269 g of Pb

8 0

3 years ago

Read 2 more answers

A student investigates the properties of an unknown liquid substance in a beaker. The student finds the substance has a mass of

shtirl [24]

Answer:

The substance is a mixture because it left a white powder in the beaker when boiled.

Explanation:

8 0

3 years ago

The ionization energy is the energy needed to remove an electron from an atom. In the bohr model of the hydrogen atom, this mean

Salsk061 [2.6K]

The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.

The first ionization energy of hydrogen is 1312.0 kJ·mol⁻¹.

Thus, H atoms in the n = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when n = ∞.

According to Bohr, Eₙ = k/n².

If n = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.

If n = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹

∴ The ionization energy from n = 2 is 328 kJ·mol⁻¹ .

6 0

3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

7 0

2 years ago