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Nata [24]
2 years ago
8

A unit of matter which cannot be broken down into other substances by ordinary chemical methods is:

Chemistry
1 answer:
Shkiper50 [21]2 years ago
6 0
a. ELEMENT: A substance<span> that </span>cannot<span> be decomposed (</span>broken down<span>) </span>into simpler substances by ordinary chemical<span> means.</span>
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A scientist investigates on two unknown solutions. Which is not a scientific experiment she can perform on the solutions?
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Which of the following objects does not decompose?
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B) Plastic Water Bottle
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"What quantitative measurement and qualitative observations can be made about the interactions of matter and energy?" can someon
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Quantitative measurements are numerical values, they involve amounts and units like measuring things. Qualitative observations appeal to the five senses, like what does the interaction look and sound like

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3 years ago
2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?
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Answer;

  = 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.  

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A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

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