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tatyana61 [14]
3 years ago
12

What molecules can cells break down for energy?

Chemistry
1 answer:
Orlov [11]3 years ago
6 0

Answer: I think the answer is C)

Explanation:

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How many atoms of oxygen (O) are in one formula unit of Mg(NO3)2?
miv72 [106K]
We may use the molecular formula of the compound to determine the number of oxygen atoms in one formula unit. The formula unit is:
Mg(NO₃)₂
Here, we can see that there are two nitrate ions in each mole, and each mole of nitrate ion contains three oxygen atoms. Thus, there are 6 oxygen atoms in each formula unit of magnesium nitrate.
5 0
3 years ago
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The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol. How much energy is produced by burning 0.8210 g
bazaltina [42]

The energy produced by burning : -32.92 kJ

<h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The enthalpy and heat(energy) can be formulated :

\tt \Delta H=\dfrac{Q}{n}\rightarrow n=mol

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol.

The energy released for 0.8210 g of naphthalene :

\tt Q=\Delta H\times n\\\\Q=-5139.6~kJ/mol\times \dfrac{0.8210~g}{128.17~g/mol}~\\\\Q=-32.92~kJ

3 0
3 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
Which is one way that scientist communicate the result of an experiment
arlik [135]

Answer:

wove oral visalize  

Explanation:

7 0
3 years ago
672cm x 11cm / 6.3cm^3
bezimeni [28]
117 333.333 m-1 your welco
5 0
3 years ago
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