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Ivanshal [37]
3 years ago
14

A chemistry student conducted several different procedures. Which of the following procedures best demonstrates the law of conse

rvation of mass?
O using filter paper to separate 2 grams of solid powder from 10 grams of water
O burning 2 kilograms of wood from a pine tree leaves 0.4 kilogram of ash and soot
O evaporating sugar water leaves 5 grams of sugar and turns the water into water vapor
O combining 2 grams of copper and 4 grams of sulfur to make 6 grams of copper sulfate

Chemistry
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

combining 2 grams of copper and 4 grams of sulfur to make 6 grams of copper sulfate

Explanation:

The Law of Conservation of Mass states that matter cannot be created or destroyed in chemical reactions. It posits that the mass of a substance remains unchanged after undergoing a chemical process.

Therefore, according to the question, the procedure that best demonstrates the law of conservation of mass after the chemistry student conducted different procedures is option D.

This is because, the mass of copper and sulfate did not change but merely combined to form a compound.

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Which one is the right spot because it really does not make any sense at all (in my viewpoint).
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The sun supplies about 1.0 kilowatt of energy for each square meter of surface area (1.0 kW/m^2 where a watt = 1 kJ/s) Plants pr
kaheart [24]

Answer:

0.092 %

Explanation:

The equation of the reaction can be computed as :

12CO_2_{(g)} + 11H_2O_{(l)} \to C_{12}H_{22}O_{11} + 12O_{2_(g)}

\Delta H = 5645 \ kJ

recall that; the number of moles = \dfrac{mass}{molar \ mass}

By applying the method of enthalpy of combustion for sucrose at the same time changing the time from hours to seconds, we can determine the total energy output.

i.e

=\dfrac{0.20g \ of \ sucrose }{m^2 \ 3600 \ s}\times \dfrac{1 \ mol}{342.34 \ g}\times 5.645 kJ/mol

= 9.16 \times 10^{-4} \ kJ/m^2 s

Given that the sun supplies about 1.0 kilowatt, to KJ/m² s, we have:

1.0 \dfrac{kW}{m^2 }= 1.0 \dfrac{kJ}{m^2 s}

Finally, the percentage of sunlight used to produce sucrose :

= \dfrac{9.16 \times 10^{-4} \ kJ/m^2 \ s}{1.0 \ kJ/m^2 . s} \times 100\%

= 0.092 %

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3 years ago
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