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AveGali [126]
3 years ago
5

Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. Th

is causes sodium azide (NaN3) to decompose explosively according to the following reaction. 2 NaN3(s) --> 2 Na(s) 3 N2(g) What mass in grams of NaN3(s) must be reacted in order to inflate an air bag to 79.5 L at STP
Chemistry
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

154 g

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

79.5 L × 1 mol/22.4 L = 3.55 mol

Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.

Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.37 mol × 65.01 g/mol = 154 g

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Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

8 0
3 years ago
Why water is able to stick to the side of a glass
Snezhnost [94]

Answer:

Water uses adhesive forces that allow it to stick to certain surfaces such as glass.

Explanation:

When the angle between vertical direction and the glass wall is small, surface tension is stronger and the component of gravity perpendicular to the glass wall is small. The result of this causes water to stick to the side of a glass.

Hope this helps!

3 0
3 years ago
Read 2 more answers
Someone plz help me
a_sh-v [17]

Answer:

A experimental investigation

3 0
2 years ago
What is the net ionic equation for the reaction that occurs when aqueous copper(II) sulfate is added to excess 6-molar ammonia?a
Karolina [17]

Answer:

c. 2NH₃ + 2H₂O  + Cu²⁺ → Cu(OH)₂(s) + 2NH₄⁺

Explanation:

A net ionic equation is a chemical equation that list only the species that are involved in the reaction.

The reaction of ammonia with copper(II) sulfate CuSO₄ in water is:

2NH₃ + 2H₂O  + CuSO₄ → Cu(OH)₂(s) + 2NH₄⁺ + SO₄²⁻

In an ionic equation, salts are written as ions, that means CuSO₄ must be written as Cu²⁺ + SO₄²⁻. That is:

2NH₃ + 2H₂O  + Cu²⁺ +<u> SO₄²⁻</u> → Cu(OH)₂(s) + 2NH₄⁺ + <u>SO₄²⁻</u>

As in a net ionic equation you must list only the species involved in the reaction (The underlined species don't react), the net ionic equation is:

<em>c</em>. <em>2NH₃ + 2H₂O  + Cu²⁺ → Cu(OH)₂(s) + 2NH₄⁺</em>

<em></em>

I hope it helps!

7 0
3 years ago
When magnesium reacts with hydrochloric acid, hydrogen gas is formed: 2HCl + Mg → H2 + MgCl2. What is the volume of hydrogen pro
Maksim231197 [3]

Answer:- C. 16.4 L

Solution:- The given balanced equation is:

2HCl+Mg\rightarrow H_2+MgCl_2

From this equation, there is 2:1 mol ratio between HCl and hydrogen gas. First of all we calculate the moles of hydrogen gas from given grams of HCl using stoichiometry and then the volume of hydrogen gas could be calculated using ideal gas law equation, PV = nRT.

Molar mass of HCl = 1.008 + 35.45 = 36.458 gram per mol

The calculations are shown below:

49.0gHCl(\frac{1molHCl}{36.458gHCl})(\frac{1molH_2}{2molHCl})

= 0.672molH_2

Now we will use ideal gas equation to calculate the volume.

n = 0.672 mol

T = 25 + 273 = 298 K

P = 101.3 kPa = 1 atm

R = 0.0821\frac{atm.L}{mol.K}

PV = nRT

1(V) = (0.672)(0.0821)(298)

V = 16.4 L

From calculations, 16.4 L of hydrogen gas are formed and so the correct choice is C.

7 0
3 years ago
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