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mr Goodwill [35]

3 years ago

14

Write the balanced equation for each of the following changes and identify whether heat is a product or needed to start the reac

tion (do not include 'heat' in your balanced equation, simply identify whether it is a product or reactant):
(a) the sublimation of dry ice [conversion of CO2(s) directly to CO2(g)] Heat is a product and is given off by the reaction. Heat is a reactant and is needed to start the reaction.
(b) the reaction of 1 mol of sulfur dioxide with oxygen. Heat is a reactant and is needed to start the reaction. Heat is a product and is given off by the reaction.

1 answer:

storchak [24]3 years ago

5 0

Explanation:

(a) the sublimation of dry ice [conversion of CO2(s) directly to CO2(g)]

The balanced equation is given as;

CO2(s) --> CO2(g)

This reaction is endothermic, hence; Heat is a reactant and is needed to start the reaction.

(b) the reaction of 1 mol of sulfur dioxide with oxygen.

The balanced equation is given as;

2SO2 + O2 --> 2SO3

This reaction is exothermic, hence; Heat is a product and is given off by the reaction.

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What is the range of 14 and 25

Alexxandr [17]

Range means the difference between the highest and lowest number. So subtract 25 - 14 which gives you 11.

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2 years ago

Read 2 more answers

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

<u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

<u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

<u>For b:</u>

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

<u>For process 2:</u>

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

<u>For process 3:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

<u>For process 4:</u>

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5730 years

Putting values in above equation, we get:

$k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 25) = 75 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs$

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0

3 years ago

Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?

Rufina [12.5K]

Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr

6 0

3 years ago

At 0 degrees Celsius, a gas occupies 22.4L. How hot must the gas be in celcius to reach a volume of 25.0L

NikAS [45]

Answer:

31.7 °C

Explanation:

Charles law states that for volume of a gas is directly proportional to the absolute temperature for a fixed amount of gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at second instance

temperature should be in kelvin scale

T1 - 0 °C + 273 = 273 K

substituting the values in the equation

22.4 L / 273 K = 25.0 L / T2

T2 = 304.7 K

temperature in celcius is - 304.7 K - 273 = 31.7 °C

the gas must be 31.7 °C to reach a volume of 25.0 L

7 0

3 years ago