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3 years ago

A standard soft drink can has a volume of 0.355 L. An empty can is sealed while

Chemistry

1 answer:

beks73 [17]3 years ago

8 0

Answer:

0.35 atm

Explanation:

To solve this problem, we use Boyle's Law: $P_1V_1=P_2V_2$ , where P is the pressure and V is the volume.

Here, V_1 = 0.355 L, P_1 = 1.0 atm, and V_2 = 0.125 L. So, just plug these values into the equation:

(1.0) * (0.355) = $P_2$ * (0.125) ⇒ $P_2$ ≈ 0.35 atm

Thus, the pressure is 0.35 atm.

Hope this helps!

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Calculate the density of a solid substance if a cube measuring 2.54 cm on one side has a mass of 108 g/cm3

kozerog [31]

Answer:

6.59 g/cm³

Explanation:

Step 1: Given data

Side of the cube (s): 2.54 cm

Mass of the cube (m): 108 g

Step 2: Calculate the volume of the cube

We will use the following expression.

V = s³ = (2.54 cm)³ = 16.4 cm³

Step 3: Calculate the density of the solid

Density is an intrinsic property, equal to the quotient between the mass and the volume.

ρ = m/V

ρ = 108 g / 16.4 cm³ = 6.59 g/cm³

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2 years ago

Which form of carbon is the hardest

Nataly [62]

Diamond are the hardest Pure form of carbon.

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2 years ago

8. The oxidation number of magnesium in magnesium chloride is _____.

Gennadij [26K]

-1 or +2 hope that helps!

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2 years ago

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What determines the volume of a gas?

Fittoniya [83]

Gases, unlike solids and liquids, have neither fixed volume nor shape. They are molded entirely by the container in which they are held.

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2 years ago

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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

3 years ago