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Nadya [2.5K]
3 years ago
9

How many oxygen atoms are present in 23.6 grams of aspartame?

Chemistry
1 answer:
olga55 [171]3 years ago
7 0
The formula of aspartame is <span>C14H18N2O5.
From the periodic table:
molecular mass of hydrogen = 1 grams
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of oxygen = 16 grams

This means that:
molar mass of aspartame = 14(12)+18(1)+2(14)+5(16) = 294 grams
Therefore, each 294 grams of aspartame contains 5(16) = 80 grams of oxygen. To know number of grams of oxygen in 23.6 grams of aspartame, simply use cross multiplication as follows:
mass of oxygen = (23.6x80) / 294 = 6.42 grams</span>
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Answer:

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8 0
3 years ago
In a previous problem, the empirical formula of a compound of phosphorus and oxygen was found to be p2o5. experimentation shows
agasfer [191]
The empirical formula for the compound is - P₂O₅
the empirical formula is the simplest ratio of whole numbers of components in a compound.
molecular formula is the actual ratio of components in a compound.
we have to first find the number of empirical units in the molecular formula 
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3 0
3 years ago
The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the
Lynna [10]

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

-\frac{d[A]}{dt} =kt

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

-\frac{d[A]}{dt} =kt

\Rightarrow -\frac{d[A]}{[A]}=kdt

Integrating both sides

\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

⇒c= -ln[A₀]  

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]=  0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

5 0
3 years ago
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attashe74 [19]

Answer:

Explanation:

Given:

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T2 = (2 × 400 × 293)/200 × 3

= 234400/600

= 390.67 K

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