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3 years ago

Identify the limiting reagent in the reaction mixture shown below (red = a2, blue = b2). the balanced reaction is a2+2b2→2ab2

1 answer:

6 0

With the equation <span>A₂+2B₂→2AB₂ we know that one mole of A₂ is necessary to react with two moles of B₂ and with that, we obtain two moles of the product. ( you can see this by just seeing what numbers are before the letters)

Since it's needed more quantity of B than A, B is the limited reagent.
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A student prepares a 0.47mM aqueous solution of acetic acid CH3CO2H. Calculate the fraction of acetic acid that is in the dissoc

Aliun [14]

The fraction of acetic acid that is dissociated is 0.18

Why?

The chemical equation for the dissociation of acetic acid (HAc) is the following:

HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)

To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:

$f=\frac{-Ka+\sqrt{Ka^{2} +4KaC} }{2C} = 0.18$

Have a nice day!

#LearnwithBrainly

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3 years ago

How are Ionic and Covalent Bonds are formed with examples ?

Helga [31]

Answer:Comparison of Ionic and Covalent Bonds

In an ionic bond, the atoms are bound together by the electrostatic forces in the attraction between ions of opposite charge. ... For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl. In a covalent bond, the atoms bond by sharing electrons.

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2 years ago

In general, for a gas at a constant volume, _____.

densk [106]

It is kept constant

There is the answer if it helped

5 0

2 years ago

After severe dust storms in the 1930s, practices were put in place to reduce soil erosion. Which

timurjin [86]

Answer: Farmers were paid to practice soil-conserving techniques like crop rotation and terracing

Explanation:

read about it here: https://www.pbs.org/wgbh/americanexperience/features/dust-bowl-surviving-dust-bowl/

5 0

2 years ago

Read 2 more answers

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

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2 years ago