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3 years ago

PLEASE HELP!!!!!!!!!

Chemistry

1 answer:

KonstantinChe [14]3 years ago

5 0

Answer:

Option "C" is the correct answer to the following question.

Explanation:

Given:

Pressure in an automobile tire (P) = 1.88 atm

Temperature (K) = 25°C = 273 + 25 = 298 Kelvin

New temperature (K1) = 37°C = 273 + 37 = 310 Kelvin

Find:

New pressure in an automobile tire (P1) = ?

Computation:

$\frac{P}{K}= \frac{P1}{K1} \\\\\frac{1.88}{298}= \frac{P1}{310}\\\\1.9557$

New pressure in an automobile tire (P1) = 1.9557

New pressure in an automobile tire (P1) = 1.96 (Approx)

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In earths surface or the bottom of the Ocean

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The right answer for the question that is being asked and shown above is that: "d. does not produce energy in nuclear power plants." The model most likely represents a reaction which d. does not produce energy in nuclear power plants

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2 years ago

Read 2 more answers

Balance the equation :

Sati [7]

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

Explanation:

Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- . Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.

Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.

Balancing the equation, we get:

2AlI 3( aq ) + 3Cl2 ( g ) → 2AlCl3 ( aq ) + 3 I 2 ( g )

Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.

Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.

We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.

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3 years ago

8. Calculate the number of moles of eachsubstance.a. 5.45 x 1026 particles of methane, CH4

Klio2033 [76]

ANSWER

The number of moles of methane is 905.32 moles

STEP-BY-STEP EXPLANATION:

Given information

The number of particles of methane = 5.45 x 10^26 particles

Let x represents the number of moles of methane

To calculate the number of moles, we will be using the below formula

$\text{Number of particles = number of moles x Avogadro's constant}$

Recall that, the Avogadro's constant is given as

$6.02\cdot10^{23}$ $\begin{gathered} 5.45\cdot10^{26}\text{ = x }\cdot\text{ 6.02 }\cdot10^{23} \\ \text{Divide both sides by 6.02 }\cdot10^{23} \\ x\text{ = }\frac{5.45\cdot10^{26}}{6.02\cdot10^{23}} \\ x\text{ = }\frac{5.45}{6.02}\cdot10^{26\text{ - 23}} \\ x\text{ = 0.9053 }\cdot10^3 \\ x\text{ = 905.32 moles} \end{gathered}$

Therefore, the number of moles of methane is 905.32 moles

6 0

1 year ago

A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.18 M solution of H₂A?

Tcecarenko [31]

Answer:

pH = 2.10

Explanation:

We name an acid as diprotic because it can release two protons:

H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁

HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂

We propose the mass balance:

Analytical concentration = [H₂A] + [HA⁻] + [A⁻²]

As Ka₂ is so small, we avoid the [A⁻²] so:

0.18 M = [H₂A] + [HA⁻]

But we can not avoid the HA⁻, because the Ka₁. Ka₁'s expression is:

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

We propose the charge balance:

[H₃O⁺] = [HA⁻] + [A⁻²] + [OH⁻]

As we did not consider the A⁻², we can miss the term and if

Kw = H⁺ . OH⁻

We replace Kw/H⁺ = OH⁻. So the new equation is:

[H₃O⁺] = [HA⁻] + Kw / [H₃O⁺]

The acid is so concentrated, so we can avoid the term with the Kw, so:

[H₃O⁺] = [HA⁻]

In the mass balance we would have:

0.18 M = [H₂A]

We replace at Ka₁

Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]

Ka1 . 0.18 / [H₃O⁺] = [HA⁻]

We replace at the charge balance:

[H₃O⁺] = Ka1 . 0.18 / [H₃O⁺]

[H₃O⁺]² = 3.4×10⁻⁴ . 0.18

[H₃O⁺] = √(3.4×10⁻⁴ . 0.18)

[H₃O⁺] = 7.82×10⁻³

- log [H₃O⁺] = pH → - log 7.82×10⁻³

pH = 2.10

5 0

3 years ago