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3 years ago

9

A pogo stick bounces 30 M in 10 secs what is the bouncers average speed?

2 answers:

DiKsa [7]3 years ago

7 0

I would love to help but it is 3 or is it 3 million?

AleksAgata [21]3 years ago

4 0

three i hope this helps

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Listed below are blood groups of o, a, b, and ab of randomly selected blood donors (based on data from the greater new york bloo

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The answer to this is 22, confirmed by gradpoint

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0.1005 liters is the same as: A. 0.0001005 cm3 B.0.1005 cm3 C.100.5 cm3 D.0.01005 cm3 and A. 0.01005 mL B. 0.1005 mL C. 0.000100

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Heredity is defined as

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Answer:

B.

the passage of genetic instructions from one generation to the next generation.

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3 years ago

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

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3 years ago

A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon

ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

$q_1=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_1=675J/^oC\times 29.62^oC=19993.5J$

To calculate the heat absorbed by water, we use the equation:

$q_2=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[19993.5+114690.12]J=134683.62J=134.7kJ$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol$

Hence, the enthalpy of the reaction is 269.4 kJ/mol

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3 years ago

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