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3 years ago

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What is the relationship between the atomic mass of an element and a mole of that element's atoms?

1 answer:

Amiraneli [1.4K]3 years ago

8 0

Explanation:

i think

Each ion, or atom, has a particular mass; similarly, each mole of a given pure substance also has a definite mass. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units (amu) or in grams per mole (g/mol).

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To break the oxygen-oxygen bonds in a single O2 molecule, 1 × 10–18 J of energy is required. Which of the following wavelengths

Sonbull [250]

E = hc/(lamda)

The lamda symbol is wavelength, which this site does not have. I can represent it with an "x" instead.

Plancks constant, h = 6.626×10^-32 J·s

Speed of light, c = 3.00×10^8 m/s

The energy must be greater than or equal to 1×10^-18 J

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x ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / (1×10^-18 J)

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The wavelength of light must be greater than or equal to 199 nm

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3 years ago

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3 years ago

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3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

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USE YOUR OWN WORDS. Each question must have a three sentence long answer!

Furkat [3]

Answer:

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Explanation:

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2 years ago

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