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aleksandr82 [10.1K]
3 years ago
15

Automobile airbags inflate due to the formation of nitrogen gas from the chemical reaction 2NaN3(s)⟶3N2(g)+2Na(s) Identify the n

umber of each atom in the reactants and products for this balanced reaction. Identify the number of Na atoms in the reactants and products.
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
6 0

<u>First part:</u>

2 atoms of Na

6 atoms of N

<u>Second part:</u>

2 atoms of N

2 atoms of Na

Why?

We are given the chemical reaction, so, we can extract the number of atoms for both reactants and products.

We are given the chemical equation:

2NaN_{3}=3N_{2}+2Na

We can rewrite it as follow:

2(NaN_{3})=3N_{2}+2Na

From the reactants, we have:

2(NaN_{3})

2Na=2atoms\\2N_{3}=2*3=6atoms

From the products we have:

3N_{2}+2Na

3N_{2}=3*2=6atoms

2Na=2atoms

Hence, we have that:

There are <u>6 atoms of N in the reactants</u> and <u>6 atoms of N in the products.</u>

and

There are <u>2 atoms of Na in the reactants</u> and<u> </u><u>2 atoms of Na in the products</u> since we have a balanced equation.

Have a nice day!

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Every afternoon, Bill and Anita go down to the beach to fly their kite. They noticed that the wind always blows towards shore. W
EastWind [94]

Answer:

the answer is D

Explanation:

Air over the beach heats up, rises, and is replaced by cool, denser ocean air

The ocean heats up much slower and becomes relatively cooler than the beach making it a high pressure zone. The air moves from the water to the land each morning forming a sea breeze. These pressure zones might equal out later in the day and the winds may diminish. ... A sea breeze blows from the ocean towards the beach.

8 0
3 years ago
Binary compound of oxygen and an unknown element,X, has the formula XO2 and is 69.55 mass % oxygen, what is the weight of elemen
marissa [1.9K]

Answer:

Weight of element X = 14.01 gram

Explanation:

Mass percentage is calculated by using :

Percent =\frac{Mass\ of\ element}{Total\ mass}\times 100

O\ Percent=\frac{Mass\ of\ O}{Total\ mass\ of\ XO2}\times 100

Mass of O = 16.0 gram

let mass of X = X grams = ?

Mass of XO2 = mass of X + 2(mass of O)

Mass of XO2 = X + 2(16)

Mass of XO2 = X + 32

Total mass of O in the compound = 2(16) = 32 grams

Percent O = 69.55 %

Percent\ O =\frac{Mass\ of\ O}{Total\ mass\ of\ XO2}\times 100

69.55=\frac{32}{X+32}\times 100

\frac{69.55}{100}=\frac{32}{X+32}

0.6955=\frac{32}{X+32}

0.6955\times (X+32)=32

0.6955X+ 22.256=32

0.6955X=9.744

X=\frac{9.744}{0.6955}

X =14.01

weight of element X = 14.01 gram

14.01 is mass of nitrogen N

The correct formula of this element is NO2

6 0
3 years ago
When 3243. grams of iron (III) chloride are reacted with 511.8 grams of hydrosulfuric acid, which is the limiting reactant?
AleksAgata [21]

Answer:

Hydrosulfuric acid will act as limiting reactant.

Explanation:

Given data:

Mass of iron(III) chloride = 3243.0 g

Mass of hydrosulfuric acid = 511.8 g

Limiting reactant = ?

Solution:

Chemical equation:

2FeCl₃ + 3H₂S       →       Fe₂S₃ + 6HCl

Number of moles of iron(III) chloride:

Number of moles = mass/molar mass

Number of moles = 3243.0 g/ 162.2 g/mol

Number of moles = 20 mol

Number of moles of hydrosulfuric acid:

Number of moles = mass/molar mass

Number of moles = 511.8 g/ 34.1 g/mol

Number of moles = 15 mol

Now we will compare the moles of both reactant with products

                      FeCl₃          :          Fe₂S₃

                       2                :            1

                      20               :          1/2 ×20 = 10

                      FeCl₃          :            HCl

                       2                :              6

                      20               :          6/2 ×20 = 60

                      H₂S             :          Fe₂S₃

                       3                :            1

                      15               :          1/3 ×15 = 5

                      H₂S            :            HCl

                       3                :              6

                      15                :          6/3 ×15 = 30

Hydrosulfuric acid producing less number of moles of product thus, it will act as limiting reactant.

 

5 0
3 years ago
The half life of Radon-222 is 3.8 days. If you have a 90.0g sample of Radon-222 in the lab, how much will be left after 19 days?
statuscvo [17]

Answer:

half lives passed=5

given sample=90g

sample left=2.8125g

Explanation:

no. of half lives=total time/half life

no.=19days/3.8days

no.=5 days

after 5 half lives sample left=2.8125g

5 0
3 years ago
How many grams of Nacl would you need to prepare 200 ml of a 5 M SOLUTION. ?
vazorg [7]

Answer:

58.44 g of NaCl are needed.

Explanation:

Given data:

Mass of NaCl needed = ?

Volume of solution = 200 mL (200/1000 =0.2 L)

Molarity of solution = 5 M

Solution:

We will solve this problem through molarity formula.

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will put the values.

5 M = moles of solute / 0.2 L

Moles of solute = 5 mol/L × 0.2 L

Moles of solute = 1 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1 mol × 58.44 g/mol

Mass = 58.44 g

Thus, 58.44 g of NaCl needed.

8 0
3 years ago
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