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3 years ago

In general, what happens to the rate of reaction when you decrease the

Chemistry

1 answer:

GaryK [48]3 years ago

3 0

Answer:

It decreases

Explanation:

Temperature change has a considerable effect on reaction rates since temperature is directly proportional to the average kinetic energy of reacting particles.

Since temperature is directly proportional to reaction rates, if the temperature is increased, the reaction rate also increases.

If the temperature rate decreases, the reaction rate also decreases.

Generally for every 10°C rise in temperature above room temperature, it has been found that reaction rates become double or triple.

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Oxidation of methane

Mandarinka [93]

<h2><em><u>ᎪꪀsωꫀᏒ</u></em></h2>

➪Methane (CH4) is oxidized with molecu- lar oxygen (O2) to carbon dioxide (CO2).

4 0

3 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

1. How much 6.0 M HNO3 is needed to neutralize 39 mL of 2.0 M KOH?

quester [9]

Answer:

For 2. the answer is 15.0 mL

For other examples, you can solve by exact way as I have solved the 2nd example.

I have writen down all the balanced chemical reaction equation for examples 1, 3, 4, 5 for you. ( picture 2 )

Explanation:

Please see the step-by-step solution in the picture attached below.(picture 1)

Hope this answer can help you. Have a nice day!

6 0

3 years ago

A 298 lb person must receive Heparin, determine the number of units given every hour (Heparin 8.0 units/kg per hour). Enter your

yarga [219]

Answer:

1,081.1 units of heparin should be given to 298 lb person.

Explanation:

We are given:

Weight of the person = 298 lb = 135.143 kg

Conversion factor used:

1 lb = 0.4535 kg

Number of unit of heparin to be given in an hour = 8.0 units/kg

Number of units given to the patient weighing 135.143 kg :

$8.0 units/kg\times 135.143 kg=1,081.144 units\approx 1,081.1 units$

1,081.1 units of heparin should be given to 298 lb person.

4 0

2 years ago

Which product formed when MgO and H2O react together?

seropon [69]

When MgO and H2O react they form Magnesium Hydroxide Mg(OH)2

4 0

3 years ago

Read 2 more answers