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Sergio [31]
3 years ago
11

Will give brainlist!!!

Chemistry
1 answer:
alina1380 [7]3 years ago
8 0

Answer:

C

Explanation:

75 mile shallow flat area just off coastlines

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If heat is added to a boiling liquid, what happens to the temperature of the liquid?
stealth61 [152]

Answer: C

Explanation:

Answer is "It does not change"

Hope this helps!

4 0
3 years ago
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
mina [271]

<u>Answer:</u> The amount of energy released per gram of B_5H_9 is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]

Taking the standard enthalpy of formation:

\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of B_5H_9 produces -9078.57 kJ of energy.

Or,

If (2\times 63.12)g of B_5H_9 produces -9078.57 kJ of energy

Then, 1 gram of B_5H_9 will produce = \frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ of energy.

Hence, the amount of energy released per gram of B_5H_9 is -71.92 kJ

8 0
3 years ago
How can you figure out how much energy your body gets from the food you eat and drink in a day?
noname [10]
It gets all the energy from food food is every important
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When 2 grams of powdered lead (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Lead (IV) ox
andre [41]

Answer:

Replacing the powdered lead oxide with large crystals

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The large crystals have less surface area exposed to the other reagents than the powdered  lead oxide. High surface area leads to a high rate of reaction thus the products are formed faster, while a low surface area leads to a lower rate of reaction since the reagent is less exposed to the other reagents.

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Melting can be best described as a process in which molecules
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Melting can be best described as a process in which molecules gain enough kinetic energy to be able to pass to each other.
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