Answer:
B: Igneous, C: Metamorphic, D: Sedimentary
Explanation:
Sorry, i dont know A
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
Answer:
The correct answer is option E.
Explanation:
Structures for the reactants and products are given in an aimage ;
Number of double bonds in oxygen gas molecule = 1
Number of double bonds in nitro dioxide gas molecule = 1
Number of single bond in in nitro dioxide gas molecule = 1
Number of triple bonds in nitrogen gas molecule = 1
![\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CNO_2%7D%5D-%5B1%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CN_2%7D-2%20mol%5Ctimes%20%5CDelta%20H_%7Bf%2CO_2%7D%5D)
(pure element)
(pure element )
The enthalpy of the given reaction is 15.86 kcal.
Answer:
i think is c,my ans will not be 100%correct
The motion of particles can be changed by the temperature you put it at.
