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2 years ago

15

If a 28.0 L balloon with a temperature of

1 answer:

Alexeev081 [22]2 years ago

6 0

Answer:

5.6 L

Explanation:

We can apply Charles' Law here since our pressure is constant (will not change inside the refrigerator) and we are relating change in volume with change in temperature:

V₁ / T₁ = V₂ / T₂ where V₁ and T₁ are initial volume and temperature, and V₂ and T₂ are final volume and temperature. Let's plug in what we know and solve for the unknown:

28.0 L / 25 °C = V₂ / 5 °C => V₂ = 5.6 L

5.6 L is our new volume (at 5 °C).

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The ratio of the lengths of two sides of a right triangle forms a

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Answer:

perimeter

Explanation:

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Which one is more metallic? <br><br> A) Gold or B) Bismuth

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Answer:

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2 years ago

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

$Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)$

The half-cell reactions are:

Oxidation half reaction (anode): $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction (cathode): $Pb^{2+}+2e^-\rightarrow Pb$

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction = ?

$[Zn^{2+}]$ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

Now put all the given values in the above equation, we get:

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

Therefore, the cell potential for this reaction is 0.50 V

5 0

2 years ago

How many chloride ions are in 2.6 moles of CaCl2?

Rama09 [41]

Answer:

31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .

Explanation:

Given data:

Number of moles of CaCl₂ = 2.6 mol

Number of Cl₂ ions = ?

Solution:

CaCl₂ → Ca²⁺ + 2Cl⁻

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

In one mole of CaCl₂ there are two moles of chloride ions present.

In 2.6 mol:

2.6×2 = 5.2 moles

1 mole Cl⁻ = 6.022 × 10²³ number of Cl⁻ ions

5.2 mol × 6.022 × 10²³ number of Cl⁻ / 1mol

31.31× 10²³ number of Cl⁻

8 0

2 years ago

What is a common land form that is formed when chemical weathering, specifically carbonation, is taking place?

skad [1K]

Answer:

plateaus

Explanation:

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2 years ago