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3 years ago

Which of these statements describes this mathematical equation?

Chemistry

1 answer:

salantis [7]3 years ago

3 0

Answer:

The final pressure of a gas is inversely proportional to the volume change and directly proportional to temperature

Explanation:

Given

$\frac{P_1V_1}{T_1} = k$

Required

Interpret

$\frac{P_1V_1}{T_1} = k$

Multiply both sides by T1

$T_1 * \frac{P_1V_1}{T_1} = k * T_1$

$P_1V_1 = kT_1$

Divide both sides by V1

$\frac{P_1V_1}{V_1} = \frac{kT_1}{V_1}$

$P_1 = \frac{kT_1}{V_1}$

This can be rewritten as:

$P_1 = k\frac{T_1}{V_1}$

In the above expression; k is a constant of proportionality.

So, the equation can be written as variation as follows:

$P_1\ \alpha\ \ \frac{T_1}{V_1}$

To interpret:

<em>P varies directly to T (the numerator) and inversely to V (the denominator).</em>

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cricket20 [7]

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3 years ago

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

2. Volume from density and mass

$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

3. Mass from density and volume

$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water =<u> 12.8 mL</u>

Volume of object = 11.8 mL

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Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

6 0

3 years ago

Find the potentials of the following electrochemical cell:

melomori [17]

Answer: 0.18 V

Explanation:-

$Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0_(Cd^{2+}/Cd)$ =-0.40V[/tex]

$E^0_(Ni^{2+}/Ni)$ =-0.24V[/tex]

$Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}$

$E^0=-0.24-(-0.40)=0.16V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}$

where,

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

$E^o_{cell}$ = standard electrode potential = 0.16 V

$E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}$

$E_{cell}=0.18V$

Thus the potential of the following electrochemical cell is 0.18 V.

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3 years ago

H+ ions increase in concentration at lower pH values. Calculate how many more H+ ions there are in a solution at a pH = 2 than i

joja [24]

The H⁺ ion concentration can be calculated from pH values using the following equation:

$pH=-log[H⁺]$

1.) Given pH = 2

Using the above equation, 2 = - log [H⁺]

Therefore, [H⁺] = 10⁻² mol/L

2.) Given pH = 6

Using the same equation, we have 6 = - log [H⁺]

Hence, [H⁺] = 10⁻⁶ mol/L

3.) Taking the ratio of [H⁺] for pH = 2 and pH = 6, we have

$\frac{10^{-2} }{10^{-6} }$ = 10⁴

So, there are 10,000 times more H⁺ ions in a solution of pH = 2 than that of pH = 6.

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3 years ago

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Which Ions oxidize aluminum?<br><br> Li^+, Ca^2+, Ag+, Sn^2+

pishuonlain [190]

Aluminum can be oxidized by $Li^{+}$ and $Ca^{2+}$

<h3>What is oxidizing?</h3>

Some of the metal corrodes (or oxidizes) and forms the corresponding metal oxide on the surface as a result of a chemical reaction between the metal surface and the oxygen in the air. The corrosion products that occur in some metals, like steel, are highly apparent and loose.

According to reactivity series (The array of metals in the descending order of their reactivities is referred to as the metals' reactivity series. It is sometimes referred to as the metals in the activity series.)

Lithium and calcium ions are more reactive than aluminum ion and they are less electronegative.

Since silver and tin are more electronegative than aluminum so, they cannot oxidize aluminum.

∴ $Li^{+}$ and $Ca^{+2}$ ions can oxidize aluminum.

Learn more about electronegativity here brainly.com/question/16446391

#SPJ10

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2 years ago