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muminat
3 years ago
10

The following pieces of evidence were found at separate explosion sites. For each item, indicate whether the explosion was more

likely caused by low or high explosive, and explain your answer.
a. Lead azide residues
b. Nitrocellulose residues
c. Ammonium nitrate residues
d. Scraps of primacord
e. Potassium chlorate residues
Chemistry
1 answer:
tangare [24]3 years ago
7 0

Explanation:

The major difference between low and high explosives is the rate of detonation. Low explosives detonate very slowly (less than 1,000 meters per second), whereas high explosives detonate very quickly (from 1,000 to 8,500 meters per second).

High explosives among the given list are Lead azide residues, Ammonium nitrate residues, and  Scraps of primacord. Whereas Nitrocellulose residues and, Potassium chlorate residues are low explosives.

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When 0.50 L of a 12 M solution is diluted to 1.0 L, what is the resulting molarity?
Lyrx [107]

Answer:

The resulting molarity is 6M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

12 M x 0,5L = C2 x 1,0 L

C2= (12 M x 0,5 L)/1,0 L

<em>C2= 6 M</em>

5 0
3 years ago
I NEED HELP ASAP!! What is the difference between the experimental group and a control group?​
dexar [7]

I believe the control group is what doesn't change in the experiment, and the experimental group is what is being tested / receives the treatment :)

3 0
3 years ago
2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)
ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}

2. Calculate the mass of KCl

m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}

3. Prepare the solution

  • Measure out 224 g of KCl.
  • Dissolve the KCl in a few hundred millilitres of distilled water.
  • Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.
8 0
3 years ago
The nucleus of a fluorine atom has a charge of? <br><br> A. +19<br> B. 0<br> C. +9<br> D. +1
posledela

Answer:

A fluorine atom has nine protons and nine electrons, so it is electrically neutral. If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.

7 0
3 years ago
Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati
worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

 

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0
3 years ago
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