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slava [35]
2 years ago
15

Explain how materials are suited for different uses based on their physical and chemical properties?

Chemistry
1 answer:
Leya [2.2K]2 years ago
4 0
Ok cool I ujust do you think I should be able to get a hold with him tomorrow morning or t if rry want me too early or something I don’t know how to get it off or if you want me too I just need it for a while and I’ll get back with him
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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
2 years ago
True or false An organism needs two dominant alleles to express that trait.
vova2212 [387]
No one can do that by him or her self

8 0
3 years ago
(a)kcl ionic (b)p4 nonpolar covalent (c)bf3 nonpolar covalent (d)so2 nonpolar covalent (e)br2 nonpolar covalent (f)no2 nonpolar
Angelina_Jolie [31]

Because K and Cl have such a large disparity in their electronegativities, KCl is a bipolar ionic molecule.

<h3>What exactly are polar and nonpolar bonds?</h3>

Polar covalent bonds develop when the distribution of electrons among atoms is uneven, whereas nonpolar side chains develop when the distribution of electrons is more even. The reason for the unequal sharing of electrons is because the atoms receiving them have various electronegativities.

<h3>How are polar bonds created?</h3>

Whenever a single pair of electrons is not shared equally, a polar molecule bond is created. This is caused by the electronegativity difference between the two elements. An unit of h as well as an unit of bromine share a pair of electrons, but not evenly.

To know more about polar bond visit:

brainly.com/question/10777799

#SPJ4

8 0
1 year ago
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
lions [1.4K]

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

4 0
3 years ago
The Thermite reaction reacts iron (III) oxide, Fe2O3 with aluminium powder, Al, to form aluminium oxide, Al2O3 and iron, Fe. Fe2
Rama09 [41]
So, I don't know the answer for b but I think I found the answer for a.

I sincerely hope this isn't wrong

3 0
3 years ago
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