Answer:
The standard enthalpy change for the reaction at
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
There are 1,000 mm in a meter so times 1.2 by 1,000 and you get 1,200.
Now since he wants it in three equal pieces divide 1,200/3 and you get 400.
So your answer is: 400 mm
Answer: a) 90.5g
b) 33.6 L
Explanation:-
Molar mass of tyrosine
= 181 g/mol
According to Avogadro's law, 1 mole of every substance weighs equal to its molar mass.
1 mole of tyrosine
weighs = 181 g/mol
0.5 moles of tyrosine
weigh 
b) According to Avogadro's law, 1 mole of an ideal gas occupies 22.4 Liters at Standard conditions of temperature and pressure (STP).
1 mole of gas at STP occupy = 22.4 L
1.5 moles of gas at STP occupy =
Answer:
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- <u>b. See the description below</u>
Explanation:
<u><em>a. Volume of 0.400 M CuSO₄(aq) required for the preparation</em></u>
In dissolutions, since the number of moles of solute is constant, the equation is:

Substitute and solve for V₁


<u><em>b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO₄(aq) from the 0.400 M CuSO₄(aq)</em></u>
You will use the stock solution, the funnel, the buret, and distilled water.
i) Using the funnel, fill in the buret with with 50 ml of the stock solution, i.e. the 100. ml of 0.400 M CuSO₄(aq) solution.
ii) Pour 37.5 ml of the stock solution from the burete into the volumetric flask.
iii) Carefully add disitlled water to the 37.5ml of the stock solution in the volumetric flask until the mark (50 ml) on the volumetric flask.
iv) Put the stopper and rotate the volumetric flask to homegenize the solution.
Answer:
Only a volcano could have produced so much ash, but there are NO volcanoes in Nebraska, where the animals were found. One Geologist in Idaho knew of a volcanic eruption dating from the same time.Bill Bonnichsen tied the 2 events together using dating techniques and comparing the ash composition from a supervolcano called Bruneau Jarbridge.