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cluponka [151]
2 years ago
15

The heat exchange in chemical reactions is due to change in

Chemistry
2 answers:
oksian1 [2.3K]2 years ago
8 0
The heat exchange in chemical reactions is due to change in enthalpy.
kirza4 [7]2 years ago
5 0
The enthalpy is the heat added to or lost in a system. 
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The percent by mass of copper in CuBr2 is
Sunny_sXe [5.5K]
Hope this helps you.

4 0
2 years ago
Read 2 more answers
How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?
Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid \rm H_2SO_4 is a diprotic acid. When one mole of \rm H_2SO_4 molecules dissolve in water, two moles of \rm H^{+} ions would be produced.

\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}.

On the other hand, sodium hydroxide \rm NaOH is a monoprotic base. When one mole of \rm NaOH formula units dissolve in water, only one mole of hydroxide ions \rm OH^{-} would be produced.

\rm NaOH \to Na^{+} + OH^{-}.

Note that \rm H^{+} and \rm OH^{-} react at a one-to-one ratio:

\rm H^{+} + OH^{-} \to H_2O.

As a result, it would take 2\; \rm mol of \rm OH^{-} to react with the \rm 2\; mol of \rm H^{+} that was released when 1\; \rm mol of \rm H_2SO_4 is dissolved in water. Since one mole of \rm NaOH formula units could produce only one mole of \rm OH^{-}, it would take \rm 2\; mol of \rm NaOH formula units to produce that 2\; \rm mol of \rm OH^{-} for reacting with 1\; \rm mol of \rm H_2SO_4.

3 0
3 years ago
Your dad is working on creating a brick border for the lake in your backyard each brick has a mass of 100 g and a volume of 20 c
wel

Answer:

5000kg/m³

Explanation:

density=mass/volume

d=m/v

d=100/20

=5g/cm³

g/cm³*1000=kg/m³

5g/m³*1000=5000kg/m³

4 0
2 years ago
Read 2 more answers
How many grams are in 3.21 x 1024 molecules of potassium hydroxide?
MakcuM [25]

Answer:

3287.04 grams

Explanation:

3 0
3 years ago
For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
2 years ago
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