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3 years ago

What is a reference point?

Chemistry

1 answer:

Kipish [7]3 years ago

6 0

Answer: A basis or standard for evaluation, assessment, or comparison; a criterion.

something that is used to judge or understand something else Definition of point of reference. : something that is used to judge or understand something else The professor used the study as a point of reference for evaluating and discussing other theories.

Explanation:

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Determine the budget for tge fitness event.consider the attendance (should tou ask for registration fee.give shirts etc)

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Notch gfi.hu,ivjmgfcmhtxumgfhkvHbyrdyu,

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3 years ago

If 5 moles of KCIO3 react completely how many moles of KCI will firm

Lubov Fominskaja [6]

With what KClO3 reacts with if it is by temperature
2KClO3===》KCl + KClO + O2

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3 years ago

Pretty Pudgey.. can i get a owaowa?

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Answer:

owaowa

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3 years ago

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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

Given that the radius of the helium-4 nucleus is approximately 2.6 fm, the classical electron radius is 2.8 fm, and the calculat

dsp73

Answer:

0.026%

Explanation:

The Helium-4 is the isotope of the helium that has a mass equal to 4. The element has 2 electrons, so, the total radius of the electrons is 2*2.8 = 5.6 fm = 0.0056 pm.

So, the total radius of the particles is 0.0056 + 0.0026 = 0.0082 pm.

The percentage of the space that the particles occupy is the radius of them divided by the radius of the atom:

% = 0.0082/31 *100%

% = 0.026%

* 1 fentometer (fm) = 0.001 picometer (pm)

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3 years ago