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2 years ago

What could you change or add to your models based on what you have learned about molecules and phases so far?

2 answers:

5 0

Answer:Phase changes require either the addition of heat energy (melting, evaporation, and sublimation) or subtraction of heat energy (condensation and freezing). ... Changing the amount of heat energy usually causes a temperature change.

Explanation:

babymother [125]2 years ago

3 0

Ummm I don’t know ‍♀️

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Which of the following is NOT a product of photosynthesis?

ryzh [129]

B is the answer..............

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3 years ago

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Calculate the energy of a photon with a frequency of 2.36 x 10-19 Hz.

OlgaM077 [116]

Answer:

Explanation:

energy for photon is calculated in same wasy as for electromagnetic radiation

energy for electromagnetic radiation = hf

where f is the frequency of photon

h is Planck's constant = h = 4.14 × 10−15 eV · s.

thus

energy of photon = 4.14 × 10−15 eV · s * 2.36 x 10-19 Hz

energy of photon = 9.77 * 10−(-15+ -19) eV

energy of photon = 9.77 * 10−34eV answer

6 0

3 years ago

Who described atoms as small spheres that could not be divided into anything smaller?

gtnhenbr [62]

John Dalton

"matter cannot be created nor destroyed or divided into smaller particles"

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3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

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3 years ago

What is a half-life in Chemistry?

makvit [3.9K]

Answer:

half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive ...

Explanation:

braniest

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2 years ago