Answer: A.Light travels in a straight line., B.Light behaves in a predictable way. , D.Light curves around corners or obstructions
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Answer:
A
Explanation:
In this question, we are to calculate the enthalpy of change of the reaction. ΔH
To be able to do that, we use the Hess’ law and it involves the subtraction of the summed heat reaction of the reactants from that of the product.
Thus, mathematically, the enthalpy of change of the reaction would be;
[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4) + 4 ΔH(Cl2)]
We can see that we multiplied some heat change by some numbers. This is corresponding to the number of moles of that compound in question in the reaction.
Also, for diatomic gases such as chlorine in the reaction above, the heat of reaction is zero.
Thus, we can have the modified equation as follows;
[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4)]
Substituting the values we have according to the question, we have;
-95.98 + 4(-92.3) -(-17.9)
= -95.98 - 369.2 + 17.9
= -447.28 KJ/mol
Answer:
See explanation
Explanation:
The mole refers to the amount of substance contained in 12 g of carbon-12. It was arbitrarily related to the number of elementary entities in 12 g of carbon -12 by Prof. Avogadro.
The number of moles of a substance is obtained as the mass of the substance divided by the mass of one mole of the substance (molar mass). Hence when the number of moles is known, the mass is now;
number of moles * molar mass
Answer:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:
<u>Answer:</u> The change in enthalpy for the given system is -642.8 kJ/mol
<u>Explanation:</u>
To calculate the change in enthalpy for given Gibbs free energy, we use the equation:
where,
= Gibbs free energy = -717.5 kJ/mol = -717500 J/mol (Conversion factor: 1 kJ = 1000 J)
= change in enthalpy = ?
T = temperature = 337 K
= change in entropy = 221.7 J/mol.K
Putting values in above equation, we get:
Hence, the change in enthalpy for the given system is -642.8 kJ/mol
