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Kobotan [32]
3 years ago
9

How many moles of aspartame are present in 2.00 mg of aspartame?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
8 0
Hope this helps, have a nice day!

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if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d
Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g

We can calculate the mass percent of O in MgO using the following expression.

\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\%  = 40.0 \%

You can learn more about mass percent here: brainly.com/question/14990953

3 0
2 years ago
Circle the letter of cach sentence that is true about silica.
miskamm [114]
The answer is either A or B I’m not sure
7 0
3 years ago
Read 2 more answers
A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found
Thepotemich [5.8K]

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

t_{1/2}=\frac{1}{k[A]_0}

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

8 0
3 years ago
Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.
melomori [17]

Answer:

0.15M

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, \frac{25.0g}{1} × \frac{1 mol}{56.106g} = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

The volume is given in L already so no need to do any conversion. M= \frac{0.4558mol}{3.00L} = 0.1485M ≈ 0.15M

5 0
2 years ago
What solute particles are present in an aqueous solution of CH3COCH3?
Anuta_ua [19.1K]
PE, GO, XY - I am  probably wrong xoxoxoxoxxo
4 0
3 years ago
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