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3 years ago

How much pressure should an airbag have?

Chemistry

2 answers:

MariettaO [177]3 years ago

7 0

Answer:

It should have about 70-80 pounds of air pressure.

slava [35]3 years ago

5 0

Answer:

5 lbs is what I read on the internetluv </3

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Iron and Oxygen form rust. Compare the mass of iron and oxygen before the reaction with the mass of rust resulting from the reac

LekaFEV [45]

The mass will stay the same because of the conservation of mass

3 0

3 years ago

Read 2 more answers

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

Answer: The amount of oxygen gas consumed by mouse is 0.202 grams.

Explanation:

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

2 years ago

The density of ethanol is 0.789 g/cm^3 at 20 degrees C. What volume of ethanol has a mass of 15.5 grams?

lys-0071 [83]

density = mass / volume

given the quotient , we have density and mass, volume can be easily calculated as:

volume = mass / p =15.5 g / 0.789 g/cm^3

=~ 20 cm^3 (dimension ally constant)

6 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

2 years ago

A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of

sergey [27]

Answer:

50 ltr 150 ltr

Explanation:

this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30

ratio of solution 1 and solution 2 =10:30

=1:3

total mixture is 200 liters

part of solution 1= $\frac{1}{4}$ ×200

=50 liters

part of solution 2= $\frac{3}{4}$ ×200

=150 liters

6 0

3 years ago