__Pb(NO3)2 + __NaCI = __NaNO3 + _

Leno4ka [110]

Answer:

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Explanation:

6 0

2 years ago

What is oxidized in a galvanic cell with aluminum and gold electrodes ?

k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

$Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V$

$Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive): $Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Notice that the overall cell potential upon summing is:

$E_{cell}=1.66 V + 1.50 V=3.16 V$

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0

2 years ago

What is the volume of 40.0 grams of argon gas at STP ?

MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up: $\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 24.9235 \ L \ Ar$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

24.9235 L Ar ≈ 24.9 L Ar

5 0

2 years ago

Two molecules that can cross a lipid bilayer without help from membrane proteins are o2 and co2. what property allows this to oc

Dennis_Churaev [7]

Two molecules that can cross a lipid bilayer without help from membrane proteins are oxygen and carbon dioxide. The property that allows this to happen is that both oxygen and carbon dioxide molecules are nonpolar which means that they can pass easily through the hydrophobic part of the membrane. The lipid bilayer is present in all cell membranes. It consists of two layers of the fat cells which are arranged into two sheets. It functions as a barrier which marks the boundaries of the cell. The inner part of a lipid bilayer is nonpolar since it is composed of the hydrophobic end of the phospholopids.

4 0

3 years ago

In a fission reaction of U-235, there was a release of 9.20 × 1011 kJ of energy. What amount of mass in kilograms would have bee

Hatshy [7]

e = mc^2

m = e / c^2

Plug in your numbers in the appropriate units.
m=1.02 × 10^-5

pulled from another response, hope it helps

3 0

2 years ago

Read 2 more answers

__Pb(NO3)2 + __NaCI = __NaNO3 + __PbCI2

Pb(NO3)2 + NaCI = NaNO3 + PbCI2