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goblinko [34]
3 years ago
9

The reported molar mass for the copper oxide formed in experiment 1 is 79.55 g/mol. what is the molecular formula of the oxide f

ormed in experiment 1
Chemistry
1 answer:
exis [7]3 years ago
4 0
Answer:
            CuO

Explanation:
                   As given the molar mass of following Copper Oxide is 79.55 g.mol⁻¹,
                                                 CuₐOₓ

Where "a" and "x" are unknown.

As we know the atomic masses of Cu and O are;

                                      Cu  =  63.55 g.mol⁻¹

                                      O  =  16 g.mol⁻¹

So, it is clear that this oxide is containing only one Copper atom,
So,
                       Molecular Mass  =  79.55
                  
                    -  Copper Mass      =  63.55
                                                    --------------
                                                       16

Hence, 16 g.mol⁻¹ is the mass of one mole of Oxygen atoms, Hence, this oxide contains only one oxygen atom.

Result:
          The molecular formula of given copper oxide is CuO.
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marusya05 [52]
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>
5 0
3 years ago
What is the pH of a solution with [H-] = 4.3 x 10-4 M?
Temka [501]

Answer:

3.37

Explanation:

pH=-log(H3O+)

- Hope that helps! Please let me know if you need further explanation.

4 0
4 years ago
A 34.57 mL sample of an unknown phosphoric acid solution is titrated with a 0.127 M sodium hydroxide solution. The equivalence p
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<u>Answer:</u> The concentration of unknown phosphoric acid solution is 0.034 M

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_3PO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?M\\V_1=34.57mL\\n_2=1\\M_2=0.127M\\V_2=28.2mL

Putting values in above equation, we get:

3\times M_1\times 34.57=1\times 0.127\times 28.2\\\\M_1=\frac{1\times 0.127\times 28.2}{3\times 34.57}=0.034M

Hence, the concentration of unknown phosphoric acid solution is 0.034 M

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3 years ago
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Explanation:

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