12 times breathe give 240 ml of pure
. Each breathe gives 20 ml of
.
Let us consider, volume of air per breathe= x ml.
Pure
from inhaled air=
ml and Pure
from exhaled air=
ml.
Pure
from inhaled and exhaled air= 20 ml
So,
+
= 20
Therefore, x = 55.5 ml
So, volume of air per breath= 55.5 ml.
Answer:
D
Explanation:
Someone's perception can be all of these things. the way one person sees something is different than the way someone standing right beside them sees it.
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
Answer:
Halogens always form anions, alkali metals and alkaline earth metals always form cations. Most other metals form cations (e.g. iron, silver, nickel), whilst most other nonmetals typically form anions (e.g. oxygen, carbon, sulfur).
Explanation:
Examples: Sodium (Na+), Iron (Fe2+), Ammonium (NH4
Explanation:
The given data is as follows.
Concentration of standard NaOH solution = 0.1922 M
Let the original acid solution concentration be x.
![x \times 25 = 250 \times M_{2}](https://tex.z-dn.net/?f=x%20%5Ctimes%2025%20%3D%20250%20%5Ctimes%20M_%7B2%7D)
![M_{2} = \frac{x \times 25}{250}](https://tex.z-dn.net/?f=M_%7B2%7D%20%3D%20%5Cfrac%7Bx%20%5Ctimes%2025%7D%7B250%7D)
= 0.1 x M
= 10.00 mL (given)
The reaction equation is as follows.
![2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O](https://tex.z-dn.net/?f=2NaOH%20%2B%20H_%7B2%7DSO_%7B4%7D%20%5Crightarrow%20Na_%7B2%7DSO_%7B4%7D%20%2B%202H_%7B2%7DO)
Concentration × Volume of
= Concentration × Volume of NaOH
![\frac{0.1 x \times 10 ml}{1} = \frac{0.1922 M \times 13.68 ml}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B0.1%20x%20%5Ctimes%2010%20ml%7D%7B1%7D%20%3D%20%5Cfrac%7B0.1922%20M%20%5Ctimes%2013.68%20ml%7D%7B2%7D)
x = 1.314 M
Therefore, we can conclude that the concentration of the original sulfuric acid solution is 1.314 M.