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steposvetlana [31]
3 years ago
12

I NEED THE ANSWER ASAP IM ON A TIMER

Chemistry
1 answer:
Ronch [10]3 years ago
6 0
That’s tuff, lol stay in school.
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Which element is the most reactive?<br> - Sodium<br> - Nickel<br> - Carbon<br> -Oxygen
hodyreva [135]

Answer:

In a reactivity series, the most reactive element is placed at the top and the least reactive element at the bottom. More reactive metals have a greater tendency to lose electrons and form positive ions .

Explanation:

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3 years ago
Cu+k.HNO3 --&gt; Cu(No3)2+..........+...........
pentagon [3]

Answer:

Cu + HNO3 → Cu(NO3)2 + <u>H2O</u> + <u>NO2</u>

Cu + 4HNO3 → Cu(NO3)2 + <u>2H2O</u> + <u>2NO2 </u>(balanced equation)

4 0
3 years ago
Which pair of elements can form an ionic compound? A) Two atoms of hydrogen B) One atom of sodium and one atom of fluorine C) On
polet [3.4K]
The answer would be B. <span>One atom of sodium and one atom of fluorine</span>
5 0
3 years ago
Read 2 more answers
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
PLEASE HELP ME ASAP!! I WILL MARK BRAINLIEST!!!
neonofarm [45]

Answer:

D

Explanation:

I did this assignment.

3 0
3 years ago
Read 2 more answers
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