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Andreyy89
3 years ago
13

To be effective, a pot holder should have low _____. viscosity conductivity malleability density

Chemistry
2 answers:
polet [3.4K]3 years ago
6 0
OK so for a pot holder to be effective it would need to have density and a low conductivity 
ollegr [7]3 years ago
4 0

Answer: Option (b) is the correct answer.

Explanation:

A pot holder should be such that when a pot is heated then its holder should not get heated because only then we will be able to hold it with our hand. Otherwise our hand will burn if pot handler will have high conductivity.

Whereas viscosity is the property of a fluid which resists its flow.

Malleability is the property of a metal in which the metal can be drawn into thin sheets.

Density is the mass per unit volume of an object.

Thus, we can conclude that to be effective, a pot holder should have low conductivity.

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PLZ HELP ILL GIVE BRAINLIST!
Schach [20]

Answer:

The energy of a wave is inversely proportional to the wavelength of the wave.

As wavelength increases, the energy of the wave decreases.

As wavelength decreases, the energy of the wave increases.

Explanation:

The energy of a wave is directly proportional to the wave's frequency. As frequency increases, so does the energy of the wave.

E\propto f (energy E is proportional to frequency f)

<u>How is this related to wavelength?</u>

Frequency is inversely proportional to wavelength. That means that as frequency increases, wavelength decreases and as frequency decreases, wavelength increases.

f\propto \frac{1}{\lambda} (frequency f is inversely proportional to wavelength \lambda)

Therefore, as wavelength increases, the energy of a wave decreases and as wavelength decreases, the energy of a wave increases.

E\propto f\propto \frac{1}{\lambda}

6 0
3 years ago
The measured quantity 0.00410 g written in scientific notation is _______. Using prefixes, the number can be written as ________
vovangra [49]

Answer:

4.10x10⁻³g.

4.10 mg.

Explanation:

Hello,

In this case, considering the given number, we can see it has two digits after the first nonzero digit that is four (4) which is also at the third decimal place, it means that the scientific notation has a negative power as it is at the left of the given zeros, therefore it is:

4.10x10⁻³g.

Moreover, since the proper prefix for the factor x10⁻³ is milli, the number with prefix is:

4.10 mg.

Best regards.

6 0
3 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
Can someone tell me the Bohr rutherford diagram of atom for calcium
Annette [7]
When the parts of the atoms are connected such as the electrons are connected around the molecules which are made up of nuetrons and protons. Calcium is the same way
6 0
3 years ago
Are these answers correct? I’m a little unsure
Naily [24]

Answer:

yes I think that they are correct

8 0
3 years ago
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