Using the following chemical reaction, perform the theoretical stoi

How much energy is transferred when 30.0 g of water is cooled from 25.0 °C to 12.7 °C.

enot [183]

Answer:

Heat transferred, Q = 1542.42 J

Explanation:

Given that,

Mass of water, m = 30 grams

Initial temperature, $T_i=25^{\circ} C$

Final temperature, $T_f=12.7^{\circ} C$

We need to find the energy transferred. The energy transferred is given by :

$Q=mc\Delta T$

c is specific heat of water, c = 4.18 J/g °C

So,

$Q=30\ g\times 4.18\ J/g-^{\circ} C\times (12.7-25)\ ^{\circ} C\\\\Q=-1542.42\ J$

So, 1542.42 J of energy is transferred.

6 0

2 years ago

Name the type of reaction involved in the conversation of ethanol to Ethan

Anestetic [448]

The conversion of ethanol to ethanoic acid is an oxidation reaction.

3 0

2 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]

Answer: The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

Explanation:

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

Please help, I will give brainylest

Anna [14]

Answer:

Explanation:

Divide both of the like fractions, then multiply the answers even if their iregular.

5 0

3 years ago

How many moles of PCl5 can be produced from 22.0 g of P4 (and excess Cl2)?

scoundrel [369]

The balanced chemical reaction is written as:

P4 + 10Cl2 = 4PCl5

To determine the moles of PCl5 produced, we use the initial amount of the limiting reactant which is P4 and the relation from the reaction. We do as follows:

22.0 g P4 ( 1 mol / 123.90 g ) ( 4 mol PCl5 / 1 mol P4 ) = 0.7103 mol PCl5 produced

8 0

3 years ago