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ioda
3 years ago
6

Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units

of the rate constant by using the units M for concentration and s for time:
Chemistry
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

Explanation:

Rate laws are mathematical expressions which define the relationship that exists between the rate of a chemical reaction and the concentration of the reacting components. Rate laws take the following form:

rate = k [A]^{x} [B]^{y} [C]^{z}...

where k = rate constant

[A], [C] and [C] represent the molar concentrations of the reactants

k, x, y, and z are determined experimentally by observing how the rate of reaction changes with the concentrations of the reactants.

If x=1, the reaction is first order for [A]. Also, if y=2, the reaction is second order for [B] and so on...

The overall order of the reaction is the sum of the orders with respect to each reactant. Therefore, the overall order of the reaction is x+y+z.

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Colligative properties depend upon the ____.
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To determine the concentration of 20.00 mL of an unknown solution of a monoprotic acid, it is titrated with 0.1093 M sodium hydr
Elanso [62]

Answer:

d) 0.1202 M

Explanation:

Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.

NaOH + HA → NaA + H₂O

The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:

22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol

The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.

The molar concentration of HA is:

2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M

7 0
3 years ago
Which of the following best explains how subsoil becomes rich in minerals?
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3 years ago
A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.58
Scilla [17]

Answer:

\% O=27.6\%

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}  =0.218molC

- Moles of hydrogen are contained in the 3.922 grams of water:

n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH

- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO

Finally, we compute the percent by mass of oxygen:

\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%

Regards.

5 0
4 years ago
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