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4 years ago

6

A gummie bear was tested through a flame-calorimeter test. the bear had a mass of 1.850 grams and the temperature of 100.0 milli

liters of water increased by 15.0 degrees celsius. how many calories were in the gummie bear? show all of your calculations.

1 answer:

Elza [17]4 years ago

6 0

Answer:- 1500 calories

Solution:- mass of bear = 1.850 g

volume of water = 100.0 mL

Density of water is 1.00 g/moL. So, mass of water would be 100.0 g.

delta T for water = 15.0 degree C

specific heat capacity for water is 1 cal/(g* degree C)

q = m x c x delta T

where, q is the heat energy, m is mass, c is specific heat capacity and delta T is change in temperature.

for water, q = 100.0 x 1 x 15.0

q = 1500 calorie

heat gained by water = heat lost by bear

So, the 1.850 g bear has 1500 cal or 1.50 Cal.

(Where, 1 Cal = 1000 cal)

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3 years ago

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

lesya692 [45]

<u>Answer:</u> The nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

<u>Explanation:</u>

For the given element $_3^6\textrm{Li}$

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

$m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu$

M = mass of nucleus = $(n_p\times m_p)+(n_n\times m_n)$

$M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu$

Calculating mass defect of the nucleus:

$\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol$

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So, $0.032694g/mol=0.032694\times 10^{-3}kg/mol$

To calculate the nuclear binding energy, we use Einstein equation, which is:

$E=\Delta mc^2$

where,

E = Nuclear binding energy = ? J/mol

$\Delta m$ = Mass defect = $0.032694\times 10^{-3}kg/mol$

c = Speed of light = $2.9979\times 10^8m/s$

Putting values in above equation, we get:

$E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol$

Hence, the nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

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4 years ago

A helium-filled weather balloon has a volume of 793 L at 16.9°C and 759 mmHg. It is released and rises to an altitude of 4.05 km

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<h3>Answer:</h3>

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<h3>Explanation:</h3>

We are given;

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But, K = °C + 273.15

Thus, initial temperature, T1 is 290.05 K
Initial pressure, P1 = 759 mmHg
New pressure at 4.05 km, P2 = 537 mmHg
New temperature at 4.05 km, T2 = 7.1 °C

= 280.25 K

Assuming we are required to calculate the new volume at the height of 4.05 km

We are going to use the combined gas law.

According to the combined gas law;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula;

$V2=\frac{P1V1T2}{P2T1}$

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3 years ago

A hydrocarbon molecule has seven carbon

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When a linear chain of carbon and hydrogen atoms contains a triple bond then it is known as alkyne.

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3 years ago

For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s)

ZanzabumX [31]

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

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Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

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As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e $Fe_2O_3$ and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

$\frac{1}{1}\times 0.3356 mol=0.3356 mol$ of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

7 0

3 years ago