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ICE Princess25 [194]
3 years ago
10

Your class goes on a field trip to observe drilling machinery. The geologists at the drill site show you displays of drilled roc

ks to compare different types of drills and methods. The drilled holes in one rock display are significantly larger than the holes drilled in the display next to it. Given this information, which statement is correct? ​
Chemistry
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

the answer is petroleum

Explanation:

This is because they need a bigger drill to get to it. They also need to go down to the level oil is found.

i hope this helps

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Draw all the structural isomers for the molecular formula C3H8O. Be careful not to draw any structures by crossing one line over
Art [367]

Answer:

The three isomers having the molecular formula C_{3} H_{8}O are drawn in the figure below.

Explanation:

5 0
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For which of the following equilibria does `"K"_("eq") = ["O"_2]`? A. O2(l) O2(g) B. 2O3(g) 3O2(g) C. 2H2O(l) 2H2(g) + O2(g) D.
Feliz [49]

For plato users

the answer is a. O2(l) O2(g)

hope this helps!

8 0
3 years ago
Read 2 more answers
1) Describe the characteristics of molecular compounds that affect their particular physical
Nitella [24]

The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.
8 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this lis
weqwewe [10]

Answer:

\Delta H_{rxn}=-2043.999kJ

Explanation:

\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]

Where n_{i} and n_{j} are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).

\Delta H_{f}^{0} is standard heat of formation and \Delta H_{rxn}^{0} is standard enthalpy change for reaction at 25^{0}\textrm{C}

So, \Delta H_{rxn}=[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}]

or, \Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]

or, \Delta H_{rxn}=-2043.999kJ

4 0
3 years ago
Please help me with this question it's due in like 3hours from now
saw5 [17]

237kj mole for both.....




4 0
3 years ago
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