Can a mini fridge still be used if you break the freezer

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

$\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g$

Therefore, the mass of $NaN_3$ required is, 166.4 grams.

3 0

3 years ago

Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to<br> neutralize it?

Sonbull [250]

In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M

The concentration of HBr is 0.127 M.

3 0

3 years ago

Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β

Fantom [35]

<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

Neptunium-237
Thallium-205
Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0

3 years ago

The heat energy needed to raise the temperature of vapor above its saturation point is called _____________________. A. sensible

lana [24]

Answer:

Option (D)

Explanation:

The super-heating is usually defined as a phenomenon where a certain amount of energy is needed to raise the temperature of the water vapor beyond its normal saturation point. This is also known as the boiling delay.

The super-heat can be mathematically written as:

Super-heat = Current temperature - Boiling point of the liquid.

Thus, super-heat refers to the amount of energy that is required to increase the temperature of vapor beyond its point of saturation.

This super-heat is essential as it helps in preventing the damages of machines like air conditioner, fridge and also helps in their soft running.

Hence, the correct answer is option (D).

7 0

4 years ago

Place the following in order of decreasing molar entropy at 298 k. H2 cl2 f2

Tanzania [10]

cl2>F2>H2

we can do this by molar mass

Hydrogen - 1

clorine - 35 x2 = 70

flourine- 18 x 2 = 36

flourine - 18

6 0

3 years ago