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Murljashka [212]
2 years ago
13

Compute Reaction Rates for All Seven Trials

Chemistry
2 answers:
jeka942 years ago
7 0

Answer : The reaction rate at 3^oC,24^oC,40^oC\text{ and }65^oC are 36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

The given formula will be,

\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}

Now we have to calculate the reaction rate at different temperatures and reaction time.

\text{Reaction rate at }3^oC=\frac{1000mg/0.200L}{138.5sec}=\frac{5000mg/L}{138.5sec}=36mg/L/sec

\text{Reaction rate at }24^oC=\frac{1000mg/0.200L}{34.2sec}=\frac{5000mg/L}{34.2sec}=142mg/L/sec

\text{Reaction rate at }40^oC=\frac{1000mg/0.200L}{26.3sec}=\frac{5000mg/L}{26.3sec}=190mg/L/sec

\text{Reaction rate at }65^oC=\frac{1000mg/0.200L}{14.2sec}=\frac{5000mg/L}{14.2sec}=352mg/L/sec

Therefore, the reaction rate at 3^oC,24^oC,40^oC\text{ and }65^oC are 36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec

Mila [183]2 years ago
5 0

Answer:

36, 146, 190, 352

Explanation:

That is the correct answer on edg

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Explanation:

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If an ionic compound were composed of a4+ and b−, which unit cell structure would give a neutral compound?
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List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)
ValentinkaMS [17]

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Explanation:

In the given chemical reaction:

2C_{10}H_{22}(l)+31O_2(g)\rightarrow 20CO_2(g)+22H_2O(g)

Reactants side = Left side

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Coefficient × Number of atoms of an element in a unit molecular formula

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Among the alkali metals, the tendency to react with other substances. A. does not vary among the members of the group. . B. incr
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A 2400.-gram sample of an aqueous solution contains 0.012 gram of nh3. What is the concentration of nh3 in the solution
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Answer : The concentration of NH_3 in the solution is, 2.94\times 10^{-4}M

Explanation :

First we have to calculate the volume of aqueous solution that is water.

Density of water = 1.00 g/mL

Mass of water = 2400 g

\text{Volume of water}=\frac{Mass}{Density}=\frac{2400g}{1.00g/mL}=2400mL

Now we have to calculate the concentration of ammonia solution.

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Formula used :

\text{Molarity}=\frac{\text{Mass of }NH_3\times 1000}{\text{Molar mass of }NH_3\times \text{Volume of solution (in mL)}}

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Therefore, the concentration of NH_3 in the solution is, 2.94\times 10^{-4}M

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