The grams of solid copper oxide must be used to prepare a solution of 0.125m concentration is 5.26 g.
According to the definition of molar concentration of a substance dissolved in a solution is defined as the ratio of the number of moles to the volume of the solution.
C = n/V
The number of moles is equal to the given mass divided by the molar mass.
n = m/Mm = n ×m
Given,
The volume of the solution of copper oxide = 0.53
Molar mass of copper oxide = 79.5
Concentration of copper oxide = 0.125
CuO = cVM
= 0.125 × 0.53 × 79.5
= 5.26g
Thus, we concluded that the grams of solid copper oxide must be used to prepare a solution of 0.125m concentration is 5.26 g.
DISCLAIMER: The above question is wrong. The correct question is
Question: In lab you have to prepare 530. 00 ml solution of 0. 125 m copper (ii) oxide. How many grams of solid copper oxide must be used to prepare a solution of this concentration?
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Answer: Significant figures are important to show the precision of your answer. This is important in science and engineering because no measuring device can make a measurement with 100% precision. Using Significant figures allows the scientist to know how precise the answer is, or how much uncertainty there is.
Explanation:
Valence electrons largely determine the chemical properties of an element.
Well, clearly the calculated value for the number of hydrating water molecules would increase above its true level, because the total weight loss would be greater than expected. This is of course undesirable, but may usually be avoided by careful application of the experimental procedures. The signs to look for include
<span>(a) loss of water of hydration usually occurs at a considerably lower temperature than decomposition of the salt, because the water molecules are not strongly bonded in the hydrated complex. Dehydration typically occurs in a broad range of temperatures, typically from 50°C to around 200°C, whereas decomposition of the dehydrated salt generally takes place at temperatures over 200°C and in some case over 1000°C. So dehydration should be performed with care - avoid over-heating the sample in order to ensure that all the water has been driven off. </span>
<span>(b) dehydration often results in a change of appearance of the sample, particularly the colour and particle size of crystalline hydrates. However, decomposition may be accompanied by an additional change at higher temperatures, which gives a warning of its occurrence. </span>
<span>(c) if it is suspected that decomposition is occurring, or that dehydration is not complete, exploratory runs of varying duration at a given temperature may be carried out. There are two criteria to judge the effectiveness of the procedure </span>
<span>(i) the weight of the sample decreases to a constant stable value: this is a sign that dehydration is complete and decomposition - which is usually a much slower process - is not occurring. </span>
<span>(ii) the calculated number of molecules of water lost should take an integer value. If it differs by more than, say, 0.1 from an integer than it is probable that one of these two undesirable effects is present. Some hydrates lose water in steps through intermediate compounds with a lower level of hydration. These may provide plateaus where the weight loss is stable but dehydration is not complete. These will, in general, not provide an integer value for the number of water molecules present (because the calculation is based on the assumption that the residual sample is completely dehydrated salt).</span>
Answer:- 6984 kJ of heat is produced.
Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.
Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:
= 6984 kJ
So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.
