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Zarrin [17]
3 years ago
15

Which of these is a pure substance A.Diamond. B.Salt C.Ocean Water

Chemistry
2 answers:
Semmy [17]3 years ago
7 0
Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances which are chemical elements. All elements are pure substances.
Sveta_85 [38]3 years ago
5 0
The Answer is - A.) Diamond
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A concentration cell is constructed by using the same half-reaction for both the cathode and anode. What is the value of standar
Akimi4 [234]

Solution :

A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.

We know,

In a standard cell,

the reduction half cell reaction is :

$Ag^+(aq)+e^- \rightarrow Ag(s) E^0 = -0.80 \ V$

The oxidation half ell reaction :

$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$

Thus the complete reaction of the cell is :

$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$

$E^0 $ cell = $E_R - E_L = 0.00  \ \text{volts}$

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2 years ago
What would be a simple way to block erosion
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7 0
2 years ago
Which element is most reactive sodium,magnesium, or aluminum and why?
ludmilkaskok [199]
Magnesium would be more reactive.
6 0
3 years ago
if 11.74 liters of gas at STP are pumped into a basketball, how many moles of gas are in the basketball? assume the basketball w
ser-zykov [4K]

Answer:

0.52 mol

Explanation:

Using the general gas equation formula:

PV = nRT

Where;

P = pressure (atm)

V = volume (Liters)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At STP (standard temperature and pressure), temperature of a gas is 273K, while its pressure is 1 atm

Using PV = nRT

n = PV/RT

n = (1 × 11.74) ÷ (0.0821 × 273)

n = 11.74 ÷ 22.41

n = 0.52 mol

There are 0.52 moles in the basketball

6 0
3 years ago
Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for
Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻     The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻   The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

<h2>[OH⁻] = 14.29 </h2>

Hope this helps

4 0
2 years ago
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