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2 years ago

Which is larger, 51.5 decigrams or 51,500 decigrams

Chemistry

1 answer:

Murrr4er [49]2 years ago

8 0

I think it's 51,500 decigrams.

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How many cm3 are there in 0.25 dm3?

aniked [119]

250cm3 i think im not sure about it though

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2 years ago

Calculate the lattice energy for NaF (s) given the following

san4es73 [151]

Answer: E

Explanation:

The lattice energy is the energy change when one mole of a crystal is formed from its components ions in its gaseous sate

Therefore lattice energy = heat of Sublimation+ ionization energy +electron affinity-(heat of formation)

Therefore lattice Energy = 109 +495 -328 +570.

Lattice energy = --923kjmol-1

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2 years ago

Which of the following sequences of elements would all have four valence electrons?

timama [110]

Answer:b

Explanation:

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2 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 $9.0\times 10^{-3}M$

3 0

2 years ago

Please help ASAP! PLEASE DO NOT GIVE ME A RANDOM GIBBERISH ANSWER OR ELSE I WILL REPORT YOU AND GIVE YOU A LOW RATING!!!

UNO [17]

Answer:

if an object weighs more than an equal volume of water, it is more dense and will sink, and if it weighs less than an equal volume of water, it is less dense and will float.

Explanation:

Hope that helps

8 0

2 years ago