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3 years ago

Consider the two functions. Which statement is true?

Chemistry

1 answer:

snow_tiger [21]3 years ago

6 0

Answer:

i think its B but i could be wrong let me know and ill fix it

Explanation:

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A 17g sample of H2O2 was decomposed to yield 1g of H2 and 16g of O2 An unknown sample containing only H and O was decomposed to

ludmilkaskok [199]

Answer:

Explanation:

Our $H_2O_2$ sample yielded 1g of $H_2$ and 16g of $O_2$ , but our unknown sample yielded 2 times as much $H_2$ for the same amount of $O_2$ .

What does this mean? that the H:O proportion for the unknown sample is twice the H:O proportion for the $H_2O_2$ sample.

What is the H:O proportion for the $H_2O_2$ sample? As we can see from its formula, it's 1:1, therefore the proportion for the unknown formula must be 2:1.

That means, two H atoms for every O atom. We could write that as: $H_2O$ and you should recognize that formula, for it is one of the most common compounds on earth, Water.

5 0

3 years ago

URGENT!! Can someone please explain this to me?

djverab [1.8K]

Answer:

$CH _{4}+ 2O _{2} → CO _{2} + 2H _{2}O$

for the balanced equation

$from \: the \: equation \\ 1 \: mole \: of \: methane \: gives \: 1 \: mole \: of \: carbon \: dioxide \\ 7.4 \: moles \: of \: carbon \: dioxide \: will \: be \: given \: by \: (7.4 \times 1)moles \: of \: methane \\ = 7.4\: moles \: of \: methane \\ \\ since \: moles \: of \: oxygen \: double \: those \: of \: methane \\ moles \: of \: oxygen \: = 7.4\times 2 \\ = 14.8 \: moles \: of \: oxygen$

4 0

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