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jeyben [28]
3 years ago
11

What is the first thing you notice about a person? .-_-

Chemistry
2 answers:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

Their smile, I don't know why my eyes just go straight to teeth

Kobotan [32]3 years ago
5 0

Answer:

How they act and if they are thre type of people I could hangout with !

Explanation:

You might be interested in
Which two transitions can magma undergo?
UNO [17]
Crystallization above the ground and crystallization below the ground
6 0
3 years ago
Read 2 more answers
if anyone has chemistry on edge and has the analyzing chemical reactions project can i please have your insta? i really need it.
Sedbober [7]

Explanation:

id -9384026088

password-1234

7 0
3 years ago
What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio
evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)

We can write the law of mass action for it:

Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change x due to the dissociation extent, we are able to rewrite it as shown below:

1.8x10^{-5}=\frac{x*x}{0.100-x}

Thus, via the quadratic equation or solve, we obtain the following solutions:

x_1=-0.00135M\\x_2=0.00133M

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0
3 years ago
Read 2 more answers
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0
3 years ago
I need help understanding this problem and the answer.
victus00 [196]

Answer:-

1440 cases

Explanation: -

We are told that 1 pallet = 45 bundles.

We are also told that 1 bundle = 32 cases.

We need to find how many cases are there in 1 pallet.

1 pallet = 45 bundles (First conversion factor)

= 45 x 32 cases (second conversion factor)

=1440 cases

Thus we see that 1 pallet has 1440 cases. We needed to use two conversion factors for this, first to convert pallet to bundle and second to convert bundle to cases.

5 0
3 years ago
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