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Brrunno [24]
3 years ago
11

What is the mass of a sample of iron if that sample lost 2300J of heat energy when it cooled from 80 oC to 30°C? The specific he

at of iron is 0.449 J /g °C.
Chemistry
1 answer:
irakobra [83]3 years ago
7 0
102 grams.
Equation:
Quantify of heat = mass x specific heat x difference in temperature
We have: quantity of heat : 2300J
specific heat: .449 J/g
difference in t: 80 - 30 = 50
Solve for mass: 2300 = mass x 0.449 x 50
mass = 102.449
2 sig-figs --> 102 grams
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Answer:

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Explanation:

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How many protons does the largest transition metal have in its nucleus?
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A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?
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<span>STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c.  so we have,</span>  

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P  

substitute the values into the equation,
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3 0
3 years ago
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

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

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
PLEASE HELP ME ASAP!!! <br> I need the answer to #9. If you can show work please do.
scoundrel [369]

The correct answer would be 32/16s


7 0
3 years ago
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