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nexus9112 [7]
3 years ago
7

Please help I need help

Chemistry
1 answer:
slava [35]3 years ago
7 0

Answer:

Covalent

Ionic

Ionic

Ionic

Explanation:

CO_{2}  - Covalent

Rest all of them are Ionic bonds in Nature.

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If its sodium it would have 11 electrons
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The theoretical yield of a reaction is 75.0 grams of product and the actual yield is 42.0g. what is the percent yield? 75.0 56.0
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In the following chemical equation, which are the reactants?
Airida [17]
In the following chemical equation, which are the reactants?
2HF + Mgo MgF2 + 2H20
>
A. 2HF + MgF2
B. 2H+ + Mgo
C. MgF2 + H20
D. MgO + H20
A
B
D
In the following chemical equation, which are the reactants?
2HF + Mgo MgF2 + 2H20
>
A. 2HF + MgF2
B. 2H+ + Mgo
C. MgF2 + H20
D. MgO + H20
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3 0
3 years ago
Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is
mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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 equilibrium reaction follows:

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

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

\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/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)]=-9078.57kJ/mol

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = \frac{-9078.57}{2}=-4539.28kJ

For per gram of compound:

Molar mass of B_5H_9 = 63.12 g/mole

\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

7 0
3 years ago
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