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Alecsey [184]
3 years ago
13

Consider only the experiment you conducted with 0.5 g. of lactose.

Chemistry
1 answer:
rjkz [21]3 years ago
8 0
Answer is: <span>the pH value(level) is the independent variable.</span><span>
</span>
Missing question: <span>We conducted an experiment where we added 0.5 g of lactose to 5 different test tubes all containing 5 different pH levels. What is the independent variable?
In this experiment pH level changes, so results depend on different pH values.
Mass of lactose is same during experiment,so t</span>he number of molecules of product formed per minute is the <span>dependent variable.</span>
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Predict the initial and isolated products for the reaction. The starting material is a 6 carbon chain where there is a triple bo
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Answer:

See explanation and image attached

Explanation:

This reaction is known as mercuric ion catalyzed hydration of alkynes.

The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.

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You are using a ruler to measure the width of a test tube. The edge of the test tube is between 3 cm and 4 cm marks. if centimet
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So, the answer should be 3.6 cm.

Here's a document that explains it well: https://www.auburn.wednet.edu/cms/lib03/WA01001938/Centricity/Domain/1360/1_Uncertainty.pdf

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Read 2 more answers
For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1
mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

Rate = k[AB]^2

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt

Where, [A_0] is the initial concentration  = 1.50 mol/L

[A_t] is the final concentration  = 1/3 of initial concentration = \frac{1}{3}\times 1.50\ mol/L = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

\frac{1}{0.5} = \frac{1}{1.50}+0.2t

\frac{1}{1.5}+0.2x=\frac{1}{0.5}

t = 6.66\ s

<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>

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3 years ago
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