2 AICI3 + 3 Ca - 3 CaCl2 + 2 Al

A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i

almond37 [142]

The half reaction that occurs at the Au electrode is 1.64

Explanation:

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e- ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e- ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1 0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is 1.64

6 0

2 years ago

How many significant figures does 602.060 have. what is its precision

ryzh [129]

That number has 6 significant numbers. It's precision would be 602.

Hope it helped.

4 0

2 years ago

A 52-gram sample of water that has an initial temperature of 10.0 °C absorbs 4,130 joules. If the specific heat of water is 4.18

FinnZ [79.3K]

Answer:

The final temperature of the water is 28.98 degree Celsius.

Explanation:

It is given that,

Mass of sample of water, m = 52 grams

Initial temperature, $T_i=10^{\circ}C$

Heat absorbed, $Q=4,130\ J$

The specific heat of water is $4.184\ J/(g^{\circ} C)$

We need to find the final temperature of the water. The heat absorbed is given by the formula as follows :

$Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{4130}{52\times 4.184 }+10\\\\T_f=28.98^{\circ} C$

So, the final temperature of the water is 28.98 degree Celsius.

3 0

2 years ago

Rubidium is comprised of two isotopes, one of which has a natural abundance of 72% and contains 48 neutrons in the nucleus. give

SOVA2 [1]

Since there are only two isotopes, then that means the other isotope has an abundance of: 100 - 72 = 28%. Let y be the mass of the first isotope, and x be the mass of the second. The equation would be

85.5 = 0.72y + 0.28x

Now, rubidium has 37 protons, and each proton has a mass of 1.00727647 amu. When neutral, it must also have 37 electrons in which each electron weighs 0.000548597 amu. Let n be the number of neutrons, in which each neutron weighs 1.008664 amu. The solution is as follows:

y = 37(1.00727647) + 37(0.000548597) + 48(1.008664)
y = 85.7054 amu

Then, x would be:
85.5 = 0.72(85.7054) + 0.28x
x = 84.972 amu
So,
x = 84.972 = 37(1.00727647) + 37(0.000548597) + n(1.008664)
Solving for n,
n = 47.27 ~ 47

Therefore, there are 47 neutrons in the second isotope.

6 0

2 years ago

The only way to determine the products of a reaction is to carry out the reaction. All chemical reactions can be classified as o

myrzilka [38]

Answer:

All the statements are correct but "all chemical reactions can be classified as one of the five general types".

Explanation:

Hello,

In this case, I assume you are looking for the wrong statement as long as the following ones are correct and matches with the foundations of chemical reactions:

- The only way to determine the products of a reaction is to carry out the reaction. This is clear, because after the chemical reaction is done, one identifies the products.

- Complete combustion has occurred when all the carbon in the product is in the form of carbon dioxide. This is clear due to a 100% conversion.

- A single reactant is the identifying characteristic of a decomposition reaction. All decomposition reactions have only one reactant which breaks into less complex species.

So the wrong statement is:

- All chemical reactions can be classified as one of five general types. This is wrong because there are four widely known chemical reactions: synthesis, decomposition, simple displacement and double displacement.

Best regards.

8 0

2 years ago