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SCORPION-xisa [38]
4 years ago
9

How many moles of aspartame are present in 1.00 mg of aspartame?

Chemistry
1 answer:
Diano4ka-milaya [45]4 years ago
3 0

Answer:- There are 3.40*10^-^6 moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is C_1_4H_1_8N_2O_5 and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})

= 3.40*10^-^6 moles of aspartame

So, there would be 3.40*10^-^6 moles of aspartame in 1.00 mg of it.

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Write the common names of trichloromethane and one of their uses?​
ankoles [38]

Answer:

One is chloroform, formerly used as an inhaled anesthetic during surgery, the primary use of chloroform today is in industry, where it is used as a solvent and in the production of the refrigerant freon.

8 0
3 years ago
A 5.00L reaction vessel is filled with 1.00mol of H2, 1.00mol of I2, and 2.50mol of HI. If equilibrium constant for the reaction
sladkih [1.3K]

Answer:

[I₂] ; [H₂]  = 0.067 mol/L

[HI] = 0.765 mol/L

Explanation:

This is the reaction:

                  I₂(g)    +    H₂(g)    ⇄    2HI (g)

Initially       1mol          1mol            2.5mol

I have the initially amount of each gas at first. Some amount (x), has reacted during the reaction.

                  I₂(g)    +    H₂(g)    ⇄    2HI (g)

React            x               x                  2x

In the equilibrium I have to subtract, what I had initially and the amount that has reacted, and then, as I had 2.5 mol of HI at the begining, I have to sum, the amount of the reaction. As I have to find out the concentrations, I have to /5L, which is the volume of the vessel.

Eq.           (1-x)/5         (1-x)/5         (2.5+2x)/5

Let's make the expression for Kc

Kc = [HI]² / ( [I₂] . [H₂])

129 =  ( (2.5+2x) /5 )² / ( (1-x) /5) . ( (1-x) /5) )

129 =   ( (2.5+2x) /5 )² / ( (1-x) /5)²

129 = (2.5+2x)² / 25  /  (1-x)² / 25

129 = (2.5 + 2x)² / (1-x)²

129 = 2.5² + 2 . 2.5 .2x + 4x² / 1 - 2x + x²

129 (1 - 2x + x²) = 2.5² + 2 . 2.5 .2x + 4x²

129 - 258x + 129x² = 6.25 + 10x + 4x²

129 - 6.25 -258x -10x + 129x²-4x² = 0

122.75 - 268x + 125x² = 0 (a quadratic function)

a = 125

b = -268

c = 122.75

(-b +- (√b² - 4ac)) / 2a

x₁ = 1.48

x₂ = 0.66

We take the x₂ value, cause the x₁, will get a negative concentration and that is impossible.

[I₂] =  ( 1-0.66 ) /5 = 0.067

[H₂] = ( 1-0.66 ) /5 = 0.067

[HI] = (2.5+ 2. 0.66) /5 = 0.765

5 0
3 years ago
What is Bose Einstein state of matter and their examples
Valentin [98]

Answer:

A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.

Examples - Superconductors and superfluids are the two examples of BEC.

Explanation:

8 0
3 years ago
Which of the following correctly illustrates the conservation of mass for the reaction below?
ANEK [815]
In order to apply the principle of conservation of mass to this reaction, we must prove that the mass on the left hand side is equivalent to the mass on the right hand side. In order to do this, we must first acquire the atomic masses of each substance involved. These are:
Fe - 56
O - 16

Now, we substitute these values into the equation:
4(56) + 3(2*16) = 2(56 * 2 + 16 * 3)
224 + 96 = 320
320 = 320

As we can see, the principle of conservation of mass holds true.
8 0
4 years ago
State the law of moment of force.​
Levart [38]

Explanation:

when in equilibrium condition the total sum of anti clockwise moment is equal to the sum of clock wise moment

5 0
4 years ago
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