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Vanyuwa [196]
3 years ago
8

What is c4??????? will give brainliest if right

Chemistry
1 answer:
Lena [83]3 years ago
5 0

C-4 or Composition C-4 is a common variety of the plastic explosive family known as Composition C, which uses RDX as its explosive agent. C-4 is composed of explosives, plastic binder, plasticizer to make it malleable, and usually a marker or odorizing taggant chemical.

Or

The explosive material in C-4 is cyclotrimethylene-trinitramine (C3H6N6O6), commonly called RDX (which stands for "royal demolition explosive" or "research development explosive").

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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh
Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

Molarity = M = \frac{1,22 mol solute}{lts solution}

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent  

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03

6 0
2 years ago
Which of the following could be true of two species that have a competitve ralationship in the same ecosystems Help me out pleas
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The answer is b. as a whole, the species is mutually beneficial to carry on each others traits and exist in the same ecosystem.
6 0
3 years ago
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The substance oxidized causes the other substance to be reduced and is called the:.
blsea [12.9K]

Answer:hydrogen Peroxide

Explanation:

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2 years ago
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Plz help I’ll mark brainliest
Neko [114]

Answer:

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2 years ago
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25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti
lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

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3 years ago
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