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3 years ago

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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

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

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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.

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3 years ago

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a_sh-v [17]

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2 years ago

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Karolina [17]

Answer:

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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:

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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>

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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]

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$2HCl+Mg\rightarrow H_2+MgCl_2$

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

The calculations are shown below:

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

= $0.672molH_2$

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n = 0.672 mol

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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.

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3 years ago