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amm1812
3 years ago
8

Oxygen forms a number of compounds with nitrogen, many of which are quite reactive. one such compound is 36.86 % by mass n . cal

culate the number of grams of oxygen present in a sample of this compound that contains 56.00 grams of n .
Chemistry
1 answer:
solmaris [256]3 years ago
7 0

Mass percentage is defined as the ratio of mass of the element to the total mass of the compound.

The formula of mass percentage is given by:

Mass percentage = \frac{mass  of  the  element}{total  mass  of  the  compound}\times 100                  (1)

mass of nitrogen = 56.00 grams

Let x be the total mass of the compound.

Put the given values in formula (1):

\frac{36.86}{100} =\frac{56.00 g}{x}

x= \frac{100}{36.86}\times 56 g

x=  151.92 g

total mass of the compound = 151.92 g

To calculate the mass of oxygen in grams, subtract the mass of nitrogen from the total mass of the compound.

Mass of oxygen =  151.92 g - 56 .00 g = 95.92 g

Thus, mass of oxygen in grams = 95.92 grams.

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The mass of magnesium, which has a density of 1.74 g/cm is 504.6 g.

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Mass is the quantity of matter. Mass can be calculated by multiplying density by volume.

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Density = 1.74

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Putting the value in the equation

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3 years ago
How many grams of Ca(NO3)2 are needed to make 25.0 g of a 15.0% Ca(NO3)2(aq)?
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3.75 g.

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<em></em>

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3 years ago
One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
Tpy6a [65]

Answer : The percent purity of Fe_2O_3 in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

Fe_2O3+3CO\rightarrow 2Fe+3CO_2

First we have to calculate the mass of Fe.

\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}

Molar mass of Fe = 55.8 g/mole

\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole

Now we have to calculate the moles of Fe_2O_3

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of Fe_2O_3

So, 3.15\times 10^4mole of Fe produced from \frac{3.15\times 10^4}{2}=15750 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3

Molar mass of Fe_2O_3 = 159.69 g/mole

\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg

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Mass of original sample = 2.86\times 10^3kg

\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100

\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%

Therefore, the percent purity of Fe_2O_3 in the original sample is 87.94 %

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