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Oxana [17]
2 years ago
7

How can I solve this question.​

Chemistry
1 answer:
asambeis [7]2 years ago
4 0
You time the atomic mass by 1 mole then divide
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Which of the following statements is generally TRUE?
Vesnalui [34]

The correct answer is option d, that is, the solubility of a solid is highly dependent on temperature.  

Solubility refers to the maximum amount of a component, which will get dissolved in a given concentration of solvent at a particular temperature. The temperature influences the solubility of both gases and solids. The temperature has a direct influence on solubility.  

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3 years ago
In some regions of the world, forests are cut down and then burned to create room for farmland. How might this local environment
UkoKoshka [18]

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6 0
2 years ago
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What is the importance of "Sodium" for human health?​
Aneli [31]

Answer:

Explanation:

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6 0
3 years ago
A 7.83 g sample of hcn contains 0.290 g of h and 4.06 g of n. find the mass of carbon in a sample of hcn with a mass of 3.37 g.
xeze [42]
To obey the Law of Conservation of Mass, the sum of all individual elements of a compound is equal to the mass of the compound. So, if HCN has a mass of 7.83 grams, then

7.83 g = mass of H + mass of C + mass of N

We know the masses of H and N to be 0.290 g and 4.06 g, respectively. Hence, we can find for the mass of C:

7.83 = 0.29 + mass of C + 4.06
mass of C = 3.48 g

As an extension to the Law of Conservation of Mass, there is also a Law of Definite Proportions. According to Dalton's atomic theory, a compound is formed from a fixed ratio of its individual elements. From our previous calculations, we know that the mass ratio of H to C to N is 0.29 g: 3.48 g:4.06 grams. The ratio could also be expressed in percentages. Let's find the mass percentage of Carbon in HCN to be used later:

mass % of Carbon = (3.48 g/7.83 g)*100
mass % of Carbon = 44.44%

So, if you collect a different mass of HCN, say 3.37 g, the corresponding mass of Carbon is equal to:

Mass of Carbon = (3.37)(44.44%)
Mass of Carbon = 1.498 g
3 0
3 years ago
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