Answer:
Hello! <em>Jaesuk~Sakai Here!! UwU</em>
Explanation:
<em>my fav cod map, is the one with the deck near the ocean ^^</em>
<em>ps: u do know which one i'm talking about, right? </em>
<em />
<em>pls help me out if u know wat map i'm talking about so i can edit this answer for u ^^</em>
This example is an controversial matter because melting an candle is considered as both physical change and chemical change because no new substances were made by melting an candle. The final answer is C.
Kinetic energy is energy in motion. B, a rolling ball would be your answer because a rolling ball is energy that is moving. The rest of the answers are wrong because the actions do not use kinetic energy.
<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
Answer:
(a) 13.7 g.
(b) 28.91 g.
Explanation:
- molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.
∴ m = (no. of moles of solute)/(mass of water (kg))
<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>
<em />
<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>
∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).
m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).
∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.
<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>
∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).
m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).
∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.
