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Verdich [7]
4 years ago
14

What is gidds free energy

Chemistry
1 answer:
sweet-ann [11.9K]4 years ago
6 0
<span>The Gibbs free energy is the system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.</span>
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Sự sắp xếp nguyên tử trong vật chất
AlexFokin [52]

Answer:

sosksjsjjs

Explanation:

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5 0
3 years ago
If 27 ml of 6.0 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?
katrin [286]

The minimum mass of NaHCO3 that  must be added to the spill to neutralize the acid is 27.216 grams


<h3>calculation </h3>

write the balanced chemical equation

2NaHCO3 +H2SO4 → Na2SO4  +2H2O +2CO2

find the moles of H2SO4 = molarity x volume in liters

volume in liters =  27/1000=0.027 l

moles is therefore= 0.027 x6=0.162 moles

by use of mole ratio of  NaHCO3: H2SO4  which is 2:1 the moles of NaHCO3=0.162 x2=0.324 moles


mass of NaHCO3= moles of NaHCO3 x molar mass of NaHCO3(84g/mol)

= 84g/mol x 0.324=27.216 grams

8 0
4 years ago
Read 2 more answers
A mixture has particles that Cannot be seen but do reflect light. Should be classified as a (homogeneous mixture, suspension, so
Digiron [165]
The answer you are looking for would be Colloid, because in colloid mixtures, the particles are big enough to reflect them in a way. Colloids can be distinguished from solutions using the Tyndall effect. Light passing through a colloidal dispersion, such as smoke or foggy air, will be reflected by the larger particles and the light beam will be visible. <span />
5 0
4 years ago
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Which atom has the largest number of neutrons
chubhunter [2.5K]
Livermorium and tennessine each of these atoms contain 177 neutrons
6 0
3 years ago
Integrated rate law for second order unimolecular irreversible
kirill115 [55]

Answer:

The rate law for second order unimolecular irreversible reaction is

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}

rearranging the ecuation

-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}

Integrating between times 0 to <em>t </em>and between the concentrations of [A]_{0} to <em>[A].</em>

\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}

Solving the integral

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

5 0
3 years ago
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