Ask question

Ask question

All categories

3 years ago

9

What does the letters say acnsrl spgei clue: a tool used to measure force!

1 answer:

DaniilM [7]3 years ago

7 0

The answer is spring scale

You might be interested in

A container of gas is initially at 0.500 atm and 25 degrees Celsius, what will be the pressure at 125 degrees Celsius?

vekshin1

0.688 would be the correct answer!

3 0

2 years ago

Referring to an activity series, which of the following combinations of reactants would not produce a successful single-replacem

loris [4]

I believe the answer is Ca + MgSO4

8 0

3 years ago

Read 2 more answers

What does osmosis depend on? A. the identity of the solute and solvent B. how much solvent is present in the solution C. the num

KengaRu [80]

Osmosis is a special kind of diffusion where solvent particles move through a semi permeable membrane from low concentration of solute to high concentration of solute.

so it depends upon

a) how much solvent is present : More the solvent on one side of semipermeable membrane more the movement of solvent particles on the other side of membrane

6 0

2 years ago

How many moles of silver are in 8.46E24 atoms of silver?

KonstantinChe [14]

Answer:

14.048 moles I believe.

Explanation:

(8.46 * 10^24) / ( 6.022 * 10^23) = 14.048

6 0

2 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

<u>Answer:</u> The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

<u>Explanation:</u>

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

<u>Reduction half reaction:</u> $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

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:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago