Answer:
The enthalpy change for the given reaction is 424 kJ.
Explanation:
We have :
Enthalpy changes of formation of following s:
(standard state)
![\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5B%5CDelta%20H_f%28product%29%5D-%5Csum%20%5B%5CDelta%20H_f%28reactant%29%5D)
The equation for the enthalpy change of the given reaction is:
=
=
The enthalpy change for the given reaction is 424 kJ.
Answer:
answer is-neutrons
Explanation:
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It depends on the context iorn is a transition metal so it can hold a charge from 1-8
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
0.4 moles
Explanation:
To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.
40gCaCO3* 1mol CaCO3/100.09gCaCO3 = 0.399640 mol CaCO3
This rounds to 0.4 moles CaCO3
