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4 years ago

How is a mixture different from a chemical reaction?

Chemistry

1 answer:

GrogVix [38]4 years ago

5 0

Answer:

A mixture is made when two or more substances are combined, but they are not combined chemically. A chemical reaction is a transformation from one set of chemicals into another set.

Explanation:

A mixture can be homogeneous or heterogeneous, when homo, you cannot see each variable but hey are there. Hetero, means you can see each variable in the mixture. It is considered a mixture and not a chemical reaction because it can be reversed. It will always be able to go back to its separate forms and keep its original composition. When there is a chemical reaction, it is hard to separate and go back to the original variables because there is a molecular bond.

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A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

When the ionic compound ki is dissolved in water what attraction pulls i– ions into solution?

weeeeeb [17]

the attraction to the positive end of a water molecule.hope this helps you

4 0

3 years ago

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How is burning magnesium similar to burning methane?

IrinaVladis [17]

They both melt according to my head

3 0

3 years ago

Blank occurs when water molecules at the surface of a pool have enough kinetic energy that the attractive forces holding them to

Hunter-Best [27]

Evaporation occurs when water molecules at the surface of a pool have enough kinetic energy that the attractive forces holding them to the liquid can be overcome

6 0

3 years ago