ANSWER FAST PLZ!!!!!!!!

Why do we use scientific notation or scientific exponential numbers in math calculations?

seraphim [82]

I believe the answer is: in order not to write very big or very small number values

6 0

3 years ago

Consider the reaction that occurs when copper is added to nitric acid. Upper C u (s) plus 4 upper H upper N upper O subscript 3

Nitella [24]

Answer:

Cu

Explanation:

In the given reaction of the addition of copper to nitric acid,

Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Cu or copper would be characterized as the reducing agent in this reaction. It is the chemical substance that is losing electrons and being oxidized due to this reduction/loss in this redox reaction as it is the metal that loses electrons by reacting with the non-metals.

8 0

3 years ago

Which of the following is the main difference between science and pseudoscience?

almond37 [142]

It should be the third one. "Science deals with facts, pseudoscience deals with theories"

4 0

3 years ago

Read 2 more answers

100 POINTS!!!

Paul [167]

what do u need help with u pls respond quickly

6 0

2 years ago

Read 2 more answers

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 %

3 0

3 years ago