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Degger [83]
2 years ago
10

Select all the correct answers.

Chemistry
1 answer:
andrey2020 [161]2 years ago
3 0

The  properties which keep the water temperature from changing much are;

  • water's high specific heat capacity
  • the large mass of water
<h3>What is specific heat capacity?</h3>

The specific heat capacity is the property of a substance that shows how much its temperature changes when it is exposed to heat.

Thus, the  properties which keep the water temperature from changing much are;

  • water's high specific heat capacity
  • the large mass of water

Missing parts:

A red-hot iron nail is immersed in a large bucket of water. Although the nail cools down sufficiently to be held bare-handed, the temperature of the water barely increases. Which properties keep the water temperature from changing much?

A.) water's high heat conductivity

B.) water's high specific heat capacity

C.) the iron nail's high heat conductivity

D.) the large mass of water

E.) the iron nail's high specific heat capacity

Learn more about heat capacity:brainly.com/question/12244241

#SPJ1

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Murljashka [212]
There’s nothing to answer
7 0
3 years ago
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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 do you balance these two chemical equations?
Vlada [557]
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KCl+F₂→KF+Cl₂
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</span>notice how both sides of the reaction have equal numbers of each atom.

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the above reaction has 2 atoms of H on the products side and only 1 atom of H on the reactants side.  That means you have to multiply HCl by 2.  Now you have the equation Mg+2HCl→MgCl₂+H₂. As you can see now we have equal numbers of all the atoms on both sides which means that that is the balanced equation.

I hope this helps.  Let me know if you have any further questions or need anything to be clarified.
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ladessa [460]

Answer:

20 degree and 4000000N

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Which statement is true of a rock layer that had another rock layer deposited on top of it?​
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