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4 years ago

The specific gravity of H2SO4 is 1.84, its density would be?

Chemistry

1 answer:

Vika [28.1K]4 years ago

4 0

The density would be 1.84 g/cm<span>³</span>

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3 years ago

It is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. Explain why there wi

REY [17]

Answer:

$E_{cell}$ = +ve, reaction is spontaneous

$E_{cell}$ = -ve, reaction is non spontaneous

$E_{cell}$ = 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

$Cu+ZnSO_4\rightarrow Zn+CuSO_4$

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}$

$E^0=-0.76-(+0.34)=-1.10V$

Thus as $E_{cell}$ is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

$Zn+CuSO_4\rightarrow Cu+ZnSO_4$

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=+0.34-(-0.76)=+1.10V$

Thus as $E_{cell}$ is positive , the reaction is spontaneous.

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4 years ago

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

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Specific heat is the amount of heat absorb or released by a substance to change the temperature to one degree Celsius. To determine the specific heat, we use the expression for the heat absorbed by the system. Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)

By substituting the given values, we can calculate for C which is the specific heat of the material.

2510 J = .158 kg ( 1000 g / 1 kg) (C) ( 61.0 - 32.0 °C)
C = 0.5478 J / g °C

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3 years ago

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