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3 years ago

If an automobile air bag has a volume of 11.7 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact

Chemistry

1 answer:

Usimov [2.4K]3 years ago

5 0

Answer:

33.95 grams of NaN3

Explanation:

Number of moles of NaN3 = mass (m)/MW = m/65 mole

I mole of NaN3 requires 22.4L air bag

m/65 mole of NaN3 required 11.7L

22.4m/65 = 11.7

22.4m = 65×11.7

22.4m = 760.5

m = 760.5/22.4 = 33.95grams of NaN3

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Frequency of 6.98 x 1013

MAVERICK [17]

Answer:7,070.74

Explanation:because frequency has a lot of energy

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3 years ago

Provide 3 examples of substances that are mixtures and list the substances in them

Vedmedyk [2.9K]

Answer:

Water and salt = salt water

Oxygen and water = sea foam

Smoke and fog = smog

Explanation:

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3 0

3 years ago

Read 2 more answers

How many grams of FeCl 3 are in 250. mL of a 0.100 M solution?

Naddika [18.5K]

Answer:

The correct answer is option B

Explanation:

$$Molarity=\frac{Weight}{Molecular \,weight} \frac{1000}{V(in \, ml)}$

Given values,

Molarity of $FeCl_3=0.100M$

Volume of solution, $V=250ml$

Molecular weight of $FeCl_3=162.2$

Substituting this values in Molarity formula, we get

$0.1=\frac{weight}{162.2} \times\frac{1000}{250} $\\$\Rightarrow 16.22=weight\times4$\\$\Rightarrow weight=\frac{16.22}{4} $\\$\therefore weight=4.06g$

5 0

3 years ago

Calculate the atomic mass of Carbon if the two common isotopes of carbon have masses of

Mars2501 [29]

Answer:

Average atomic mass of carbon = 12.01 amu.

Explanation:

Given data:

Abundance of C¹² = 98.89%

Abundance of C¹³ = 1.11%

Atomic mass of C¹² = 12.000 amu

Atomic mass of C¹³ = 13.003 amu

Average atomic mass = ?

Solution:

Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100

Average atomic mass of carbon= 1186.68 + 14.43333 / 100

Average atomic mass of carbon = 1201.11333 / 100

Average atomic mass of carbon = 12.01 amu.

5 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago