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kondaur [170]
2 years ago
14

Which of the choices below is an empirical formula

Chemistry
1 answer:
Marta_Voda [28]2 years ago
5 0

Answer: It's the a12o3

Explanation: It is because I know it...

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At 1 atm, how much energy is required to heat 39.0 g of h2o(s at –24.0 °c to h2o(g at 121.0 °c?
earnstyle [38]
Q = mcΔt, q = energy [J] m = mass (of water) [g]; c = specific heat capacity of water [J g⁻¹ K⁻¹/°C⁻¹]; Δt = change in temperature [K/°C]
Δt = 121 - -24 = 145
q = 39 × 4.18 × 145
q = 23637.9 J
4 0
2 years ago
A chemical reaction in which two or more elements switch places is called?
Nina [5.8K]

Answer:

neutralization

Explanation:

8 0
3 years ago
What is the Celsius temperature of 1 mole of a gas that has an average kinetic energy of 4,290 joules?
timofeeve [1]
Well the the answer is 70.8c but if you round it up it is 71c which I choice and got it correct so the answer is 71c

5 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
12.23 g of ammonia (NH3) are dissolved in enough water to make 560.0 mL of solution. How many moles of NH3 are added to the wate
Karolina [17]

Answer:

0.719 moles of NH₃

Explanation:

Molar mass of ammonia 17 g/mol

Mass of amonia = 12.23 g

Mass / Molar mass = Moles

12.23 g / 17 g/mol = 0.719 moles

4 0
3 years ago
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