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3 years ago

What are Free Radicals ?

Chemistry

2 answers:

DENIUS [597]3 years ago

4 0

Answer:

an uncharged molecule (typically highly reactive and short-lived) having an unpaired valency electron.

larisa [96]3 years ago

4 0

<u>an uncharged molecule (typically highly reactive and short-lived) having an unpaired valency electron.</u>

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What elements don't form bonds

Agata [3.3K]

Answer:

Noble gases are a set of elements in the periodic table because they don't naturally bond with other elements. *Examples ...Helium; Neon; Radon; Xenon; Argon etc

Explanation:

theyre noble gases.

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3 years ago

Temperature is a measure of the average energy of motion of an objects particles. Sometimes, if enough heat is added to or remov

Ivan

B: a liquid becomes a gas Let me know if it’s right or wrong

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3 years ago

A graduated cylinder is filled with 60.0 ml of water. A cube, made of aluminum, is carefully dropped into the cylinder. Aluminum

katen-ka-za [31]

Answer:

65 ml

Explanation:

The aluminum will not float , so it will displace a volume of fluid equal to its volume.

13.5 gm / 2.7 gm/ml = 5 ml

the new graduated cylinder measurement will be 60 + 5 = 65 ml

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2 years ago

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maxonik [38]

The correct answer would be D) c<span>odominance, I believe.</span>

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3 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

3 years ago

What are Free Radicals ?​

What are Free Radicals ?