Answer:
The answer to your question would be substance, but chocolate power mixed into milk would be more of a suspension.
Explanation:
Neither chocolate powder nor milk are elements. They are both complex molecules. Their mixture will not result in the formation of a compound since no chemical reaction will take place.
The molecules of the chocolate powder will simply intermingle with the fatty molecules of the milk to form the substance.
When thoroughly mixed the solution will become homogeneous so there will be no lumps of chocolate power visible. But after time, the chocolate will become visible at the bottom of the clear container in which we asked you to prepare the mixture.
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Answer:
.000001585
Explanation:
14-8.20=5.8
10^-5.8 = H3O+
- Hope this helped! Please let me know if you need further explanation.
Answer and Explanation:
The resonance contributor in cyclopentadienone (as shown in the attachment below) results into the compound having a positive charge on the carbonyl group, C=O which accounts for a highly reactive anti-aromatic 4π system. And this illustrates the reason for its instability.
Answer:
increase in solvent temperature will increase the solubility of solid particle but decreases the solubility of gas particle.
Explanation:
In solid particle when temperature increases its help to break apart the solid particle and increase in kinetic energy of solvent results in increase in solubility of solid particles in solvent.
but in gas solute, increase in temperature of solvent causes the increase in motion of gas molecules means increase in kinetic energy of molecules in the gas which results in breakage of inter molecular bonds and removal of the molecules from the heated solution.
The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.
Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.
We recognise that
h * 2 = PT - P * h * 20 = 1atm - 0.037atm
= 0.963 atm
1 * h * 2 = Ph * 2V / R * T
= 0.963 atm x 0.520 L / 0.0821 L atm/
molK * 301
= 0.02 mol h2
= 0.02molZn
So 0.02 mol Zn x 65.39 g/mol
= 1.33 g Zn
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