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aleksandrvk [35]
3 years ago
12

Sodium azide is stable at room temperature but decomposes quickly at temperatures above 300°C. It is moderately inexpensive to m

anufacture but is highly toxic. Read this article on the effects of undeployed airbags. Based on the article, does sodium azide seem like a good match for the criteria and constraints you listed in part A? In what ways is sodium azide a good choice for an airbag design, and in what ways is it not?
https://app.edmentum.com/content-delivery//resource/90760c74-ecf7-42a4-aeee-3ef8fae15eab
Chemistry
1 answer:
Katen [24]3 years ago
4 0

The heat generated causes sodium azide to decompose into sodium metal and nitrogen gas, which inflates the car's air bags. Under normal circumstances, this molecule is quite stable. If heated, though, it will fall apart.Eating as little as 50 milligrams (less than two-thousandths of an ounce) of sodium azide can lead to collapse and a coma-like state within five minutes as blood pressure plummets and heart rate skyrockets. Ingest a few grams, and death occurs within 40 minutes.To prevent microbial contamination, sodium azide can be added to an antibody preparation to a final concentration of 0.02% (w/v). Many Abcam antibodies already contain this preservative at concentrations ranging from 0.02 to 0.05%. This will be indicated on the datasheets in the section titled “Storage buffer”Sodium azide is highly toxic. Ingestion of 100 to 200 mg can result in headache, respiratory distress, and diarrhea. Target organs are the central nervous system and the brain.Heavy metal azides, such as lead azide are primary high explosives detonable when heated or shaken. Heavy-metal azides are formed when solutions of sodium azide or HN3 vapors come into contact with heavy metals or their salts.

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What is the molarity of an H3PO4solution if 15.0 mL is completely neutralized by 38.5 mL of 0.150 MNaOH? Please explain your ans
rodikova [14]

Answer:

M=Molarity

V=volume

change millilitre to litre

so:15ml=0.015 L

:38.5ml=0.0385 L

M1V1=M1V1

M1* 0.015 L= 0.15M*0.0385 L

M1=0.00385

8 0
4 years ago
Substance A and B are the same mass and at the same initial temperature. When the same amount of heat is added to each, the fina
Bingel [31]
That both of them are warm....at a certain temputurature causing one to fell,cozy
7 0
3 years ago
Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re
lapo4ka [179]
a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K 

when ΔG° is Gibbs free energy 

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 =  298 K 

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

      = -7739 J  = -7.7 KJ


b) then, to get E
° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

              = - 2.7 x 10^-2 V
  
8 0
3 years ago
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
How many grams of zinc would be required to produce 9.65g of zinc hydroxide
Effectus [21]

The question is incomplete, here is the complete question:

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Zn+2MnO₂+H₂O→Zn(OH)₂+Mn₂O₃

<u>Answer:</u> The mass of zinc required is 6.35 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of zinc hydroxide = 9.65 g

Molar mass of zinc hydroxide = 99.4 g/mol

Putting values in equation 1, we get:

\text{Moles of zinc hydroxide}=\frac{9.65g}{99.4g/mol}=0.0971mol

The given chemical equation follows:

Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3

By Stoichiometry of the reaction:

1 mole of zinc hydroxide is produced from 1 mole of zinc

So, 0.0971 moles of zinc hydroxide will be produced from = \frac{1}{1}\times 0.0971=0.0971mol of zinc

Now, calculating the mass of zinc from equation 1, we get:

Molar mass of zinc = 65.4 g/mol

Moles of zinc = 0.0971 moles

Putting values in equation 1, we get:

0.0971mol=\frac{\text{Mass of zinc}}{65.4g/mol}\\\\\text{Mass of zinc}=(0.0971mol\times 65.4g/mol)=6.35g

Hence, the mass of zinc required is 6.35 grams

6 0
3 years ago
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