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2 years ago

6

Calculate the following:[6.6 x

="10^{-2}" align="absmiddle" class="latex-formula">] - [6.6 x $10^{-3}$ ]

1 answer:

anyanavicka [17]2 years ago

4 0

0.066-0.0069= 0.0591

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Which of the following elements matches the criteria below?

Tems11 [23]

Answer:

The answer to your question is below

Explanation:

1. Found in period 2. All the elements in the list are found in period 2.

a. F This option is correct

b. Be Beryllium is located in period two.

c. O also oxygen is found in period 2.

d. C Carbon is found in period 2.

2.- Can gain lose 4 electrons to become its nearest stable noble gas. Only Carbon.

a. F This option is wrong, F becomes stable when it gains 1 electron.

b. Be Beryllium becomes stable when it loses 2 electrons.

c. O Become stable when it gains 2 electrons.

<u>d. C </u><u>Become stable when it gains or loses 4 electrons.</u>

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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

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

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

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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