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2 years ago

Just vibin yall how about you

Chemistry

2 answers:

Misha Larkins [42]2 years ago

8 0

Answer:

<h2>hyy mate imm need friend I didn't have any friend can you my friend^_^</h2>

Explanation:

Inspired by the McCarthy hearings of the 1950s, Arthur Miller's play, The Crucible, focuses on the inconsistencies of the Salem witch trials and the extreme behavior that can result from dark desires and hidden agendas.

and women

Vlada [557]2 years ago

4 0

Answer:

Not much just operating brainly and waiting for my friends :)

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In a science experiment, Javi concludes that a chemical reaction has occurred. What evidence would support this conclusion?

Naily [24]

Answer:

B) a substance's color and odor changed Explanation: A signal that a chemical change has occurred is when its odor (its smell) or its appearance has changed.

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2 years ago

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Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

natka813 [3]

Answer:

Explanation:

You need to remember that the oxidation number of H is +1, except when it is in a metal hydrites like NaH, where its oxidation number is -1. Then, the oxidation number of O is -2, but in peroxides is -1. So with these rules you just have to multiply the ox. number with the name of atoms and all the elements in the reaction must sum 0.

5 0

3 years ago

Please help Please help

alexdok [17]

Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole

5 0

1 year ago

Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

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

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

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:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

4 0

2 years ago

El hipoclorito de sodio (NaClO), es vendido en una solución clara de ligero color verde-amarillento y un olor característico que

vivado [14]

Respuesta:

16,7 mL

Explicación:

Paso 1: Información provista

Concentración inicial (C₁): 6%

Volumen inicial (V₁): ?

Concentración final (C₂): 0,5%

Volumen final (V₂): 200 mL

Paso 2: Calcular el volumen de la solución concentrada

Queremos preparar una solución diluida de hipoclorito de sodio a partir de una concentrada. Podemos calcular el volumen inicial que debemos tomar usando la regla de dilución.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0,5% × 200 mL / 6% = 16,7 mL

6 0

2 years ago