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Allushta [10]
3 years ago
13

Jewelry is commonly weighed in carats. five carats are equivalent to one gram. how many grams of gold are contained in a 24-cara

t gold chain?
a. 120 g

b. 4.8 g

c. 5.0 g

d. 100 g
Chemistry
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer: b. 4.8 g

Explanation:

Given : 5 carats = 1 gram

Thus if 5 carats is equivalent to = 1 g

24 carats is equivalent to = \frac{1g}{5carat}\times 24carat=4.8g

Thus 4.8 grams of gold are contained in a 24-carat gold chain.




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Electron pair geometry and molecular geometry may be the same or may be different.
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3 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
30 pts! Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liter
mihalych1998 [28]
First, calculate the number of moles of sodium present with the given mass,

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It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.

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Then, at STP, 1 mol of gas = 22.4 L.
                
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Answer: 15.34 L
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The analysis of compound only magnesium, phosphorus and oxygen showed 36.23% MgO and 63.77 % P2O5. set up the simplest formula
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Answer:

dfgs

Explanation:

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