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ivanzaharov [21]
3 years ago
11

How many kilojoules are released when 2.25 mol of H2O2 reacts?

Chemistry
1 answer:
Stels [109]3 years ago
8 0

Answer:

230.3 kJ

Some chemical reactions show release of heat while some show absorption of heat. On these bases, the reaction is classified as either endothermic or exothermic. Negative sign of ΔH means exothermic reaction.

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can anyone answer these questions? this is due today so if anyone can help that’d be greatly appreciated
Olegator [25]

Answer:

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Explanation:

3 0
3 years ago
Potassium (k combines with magnesium bromide (mgbr2 to form potassium bromide (kbr and magnesium (mg during a single replacement
Agata [3.3K]
In balancing reactions, the number of atoms on each side should be of equal number. It is the most important rule in reactions. Also, we should know the correct substances involved in the reaction. We do as follows:

2K + MgBr2 = 2KBr + Mg
5 0
3 years ago
Read 2 more answers
If Marie and Calvin dissolve 50 grams of KBr in 100 grams of water at 90oC, the solution is
hram777 [196]

Answer:

The solution is 50 %wt

Explanation:

50% wt is a sort of concentration and means, that 50 g of solute (in this case, the potassium bromide) dissolved in 100 g of water.

It is the same to say, that there are 50g of KBr for every 100g of H₂O

8 0
3 years ago
The chemical equation may not be balanced with HCI + NaOH
Masja [62]

Answer:

idk

Explanation:

3 0
3 years ago
Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise
USPshnik [31]

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

                  2 N2O5(g)  --> 4 NO2 + O2

Initial:          0.110                  -             -

change:        -2x                  +4x        +x

Final:          0.110 - 2x           +4x        +x

But final = 0.150atm;

0.110 - 2x    +  4x   +  x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k =  rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3)  t

upon solving, t = 37.1 s

3 0
3 years ago
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