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Mkey [24]
2 years ago
9

7. What produces more severe burns, boiling water or steam?​

Chemistry
1 answer:
Gwar [14]2 years ago
7 0

Answer:

boiling water

Explanation:

boiling water produces more severe burns

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Using the following data, determine the standard cell potential E^o cell for the electrochemical cell constructed using the foll
Nana76 [90]

Answer: C)  0.637 V

Explanation:

The balanced redox reaction is:

Zn(s)+Pb^{2+}(aq)\rightarrow Zn^{2+}(aq)+Pb(s)

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

Zn^{2+}(aq)+2e^-\rightarrow Zn(s)= -0.763

Pb^{2+}(aq)+2e^-\rightarrow Pb(s)= -0.126

E^0_{[Zn^{2+}/Zn]}=-0.763V

E^0_{[Pb^{2+}/Pb]}=-0.126V

E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Zn^{2+}/Zn]}

E^0=-0.126-(-0.763V)=0.637V

The standard emf of a cell is 0.637 V

8 0
3 years ago
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic aci
Anton [14]

Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid</em>

<em />

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] <em>(1)</em>

<em></em>

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M  * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] <em>(2)</em>

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

<h3>15.2mL of the 0.10M sodium formate solution</h3>

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

<h3>4.8mL of the 0.10M formic acid solution</h3>
4 0
3 years ago
How would I Determine the number of moles in 3.51 x 10^23 formula units of CaCl2
Andrej [43]

Answer:

by using this formula you will get it

Explanation:

number of mole = number of particles÷ Avogadro's number

n=3.51×10^23÷ 6.02×10^23

n = 0.58 moles

8 0
2 years ago
Someone please help me real quick pls :(
Genrish500 [490]
Anti-acids because they contain Alkaline ion that chemically neutralizes stomach acid
8 0
3 years ago
Ted used a balance scale to weigh a 2.233 g sample of copper sulfate. Which of these measurements made by Ted is the most accura
Vika [28.1K]
There’s only one measurement the question doesn’t make sense
6 0
3 years ago
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