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2 years ago

What is the hydroxide ion concentration of the lake described in question 3?

1 answer:

3 0

The hydroxide concentration is written as [OH-]. A useful equation is that pH + pOH = 14. Since the pH is 4, the pOH must be 10. To get [OH-], again take the neg. anti-log of both sides: - anti log (pOH-) = - antilog(10) ---> [OH-]=1 x 10^-10 M or 1 x 10^-10 moles [OH-] per liter solution (in this case, lake).
The hydroxide ion concentration describe in question 3 is 10-10 M

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Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

1 year ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

$E_{cell} = E_{red} - E_{ox}$

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

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I hope it helps!

8 0

2 years ago

1. Copper (__) is an element on the periodic table.

user100 [1]

Answer: Cu

Explanation: It is Cu because the origin of the word Copper comes from the latin word "Cuprum".

3 0

2 years ago

Metals with ________ electron configurations characteristically form diamagnetic, square planar complexes.

erastovalidia [21]

Answer:

Explanation:

Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.

To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes

8 0

2 years ago

Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i

natka813 [3]

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL

4 0

2 years ago