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The balanced equation for a hypothetical reaction is a + 5b + 6c → 3d + 3e. what is the rate law for this reaction? rate = k [a]

nekit [7.7K]

Answer: The correct rate of the reaction is $Rate=k[a][b]^5[c]^6$

Explanation:

Rate law of the reaction is the expression which expresses the rate of the reaction in the terms of the molar concentrations of the reactants with each term raised to the power of their respective stoichiometric coefficients in a balanced chemical equation.

For the given reaction:

$a+5b+6c\rightarrow 3d+3e$

The expression for the rate law will be: $Rate=k[a][b]^5[c]^6$

7 0

3 years ago

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

Answer: The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

Explanation:

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago

Circle which of the following depend on the amount of matter present:

yan [13]

It is the boilimg point now aa

4 0

3 years ago

A gaseous mixture consisting of nitrogen, argon, and oxygen is in a 3.5-L vessel at 25C. Determine the number of moles of oxygen

Step2247 [10]

Answer:

Number of moles of oxygen = 0.037 mol

Explanation:

Given data:

Total pressure = 98.5 KPa

Partial pressure of nitrogen = 22.0 KPa

Partial pressure of argon = 50.0 KPa

Volume = 3.5 L

Temperature = 25°C (25+273= 298K)

Number of moles of oxygen = ?

Solution:

Total pressure = P(N₂) + P(O₂) + P(Ar)

98.5 KPa = 22.0 KPa +P(O₂) + 50.0 KPa

98.5 KPa = 72.0 KPa +P(O₂)

P(O₂) = 98.5 KPa - 72.0 KPa

P(O₂) = 26.5 KPa

KPa to atm:

26.5 KPa/ 101 = 0.262 atm

Number of moles of oxygen:

PV = nRT

n = PV/RT

n = 0.262 atm × 3.5 L / 0.0821 atm.L/mol.K × 298 K

n = 0.917atm.L /24.47atm.L/ mol

n = 0.037 mol

6 0

2 years ago

If you wanted to perform a controlled experiment to test the effect of temperature on the pressure of a bicycle tire, which of t

Sav [38]

It would be the needle pressure gage temp check

6 0

3 years ago