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sladkih [1.3K]
3 years ago
11

A teaspoon of the carbohydrate sucrose (common sugar0contains 16 Calories (16 KCal). What is the mass of the teaspoon of the suc

rose if the average number of Calories for the carbohydrates is 4.1 Calories/g?
Chemistry
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

3.902 grams is the mass of the teaspoon of the sucrose.

Explanation:

Calories in 1 teaspoon of carbohydrate sucrose = 16 calorie

The average number of calories for the carbohydrates is 4.1 Calories/g

Then number of grams in 1 calorie =\frac{1}{4.1} g

Mass of 16 calories:

16 caloire=16\times \frac{1}{4.1 }g=3.902 g

3.902 grams is the mass of the teaspoon of the sucrose.

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What is extrapolation?
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A. Predicting data that fall beyond a known data point

Explanation:

Extrapolating is unreliable because you are predicting data outside of the data range - anything could happen for the data to stop following the trend or pattern

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For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

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Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

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A=A_o\times e^{-\lambda t}

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\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

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\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

A=0.9715\times x

\frac{A}{A_0}=\frac{A}{x}=0.9715

0.9715 Fraction of Pu-239 will be remain after 1000 years.

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Why do electrons transition between energy levels within the atom, and how do we detect these transitions?
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Answer:

See explanation

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Electrons transition between energy levels in an atom due to gain or loss of energy. An electron may gain energy and move from its ground state to one of the accessible excited states. The electron quickly returns to ground state, emitting the energy previously absorbed as a photon of light. The wavelength of light emitted is measured using powerful spectrometers.

Atoms can be excited thermally or by irradiation with light of appropriate frequency.

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