To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:
mass moles
Sn 78.77 0.6636
O 21.23 1.3269
Dividing the number of moles of each element with the smallest value, we will have the empirical formula:
moles ratio
Sn 0.6636 0.6636/0.6636 = 1
O 1.3269 1.3269/0.6636 = 2
The empirical formula would be SnO2.
5.5/-38 / 5.5 = 1/-6.9 x 1.3 = 1.3/-8.98
unit volume/temperature x searching amount
i’d say the temperature would be -8.98 C simply - I don’t know what formula youd use for this
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Answer:True</h2>
Explanation:
Heterogeneous mixture is a mixture with non-uniform composition.
The properties of the mixture like concentration may change for different parts of the mixture.
Colloids contain solute particles of size
.The presence of these particles makes the mixture heterogeneous.
Suspensions contain solute particles of size
.These particles settle to the bottom of the mixture which makes the composition of the bottom different from the top.
So,colloids and suspensions are two types of heterogeneous mixtures.
Answer:
10 kg of ice will require more energy than the released when 1 kg of water is frozen because the heat of phase transition increases as the mass increases.
Explanation:
Hello!
In this case, since the melting phase transition occurs when the solid goes to liquid and the freezing one when the liquid goes to solid, we can infer that melting is a process which requires energy to separate the molecules and freezing is a process that releases energy to gather the molecules.
Moreover, since the required energy to melt 1 g of ice is 334 J and the released energy when 1 g of water is frozen to ice is the same 334 J, if we want to melt 10 kg of ice, a higher amount of energy well be required in comparison to the released energy when 1 kg of water freezes, which is about 334000 J for the melting of those 10 kg of ice and only 334 J for the freezing of that 1 kg of water.
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