A combustion reaction is one type of what class of reactions?

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antiseptic1488 [7]

Answer:

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3 0

2 years ago

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What is the Molarity of a solution of HNO3 if it contains 12.6 moles in a<br> 0.75 L solution? *

Korolek [52]

Answer:

M = 16.8 M

Explanation:

<u>Data:</u> HNO3

moles = 12.6 moles

solution volume = 0.75 L

Molarity is represented by the letter M and is defined as the amount of solute expressed in moles per liter of solution.

$M=\frac{moles}{solution volume}$

The data is replaced in the given equation:

$M=\frac{12.6 mol}{0.75L}=16.8\frac{mol}{L}$

7 0

3 years ago

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A new element (solid at room temperature) is discovered that has:

gtnhenbr [62]

Answer:

A metalloid is a substance that has both the qualities of a metal and a non metal. Metals conduct electricity and have a high melting point.The element described does not conduct electricity until temperatures are reduced therefore it is not a metal. The fact that conduction occurs when temperatures are reduced means that it is not a non metal because non metals do not conduct electricity at all. Therefore the element is a metaloid because it exhibits some properties of a metal and others of a non mental.

Explanation:

5 0

2 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

2 years ago

Pure substances include

eduard

B. elements and compounds

4 0

3 years ago