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N76 [4]
3 years ago
13

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

Chemistry
1 answer:
kobusy [5.1K]3 years ago
5 0

Answer:

0.2 x 10^19

Explanation:

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Consider the hydrocarbon below.
Ket [755]
It is an Alkene because it has a double bond, so it’ll have “ene” at the end. The simplest Alkene has 2 carbons.

2 carbons = “eth”

Look at that! Two carbons! It must be “ethene”
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How many of protien consumed by humans around the world is fish?
Alex17521 [72]

The answer would be 20% (B)

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How does a scientist make two solutions with the same molarity? A By dissolving the maximum amount of each substance in the B. B
aalyn [17]

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Explanation:

7 0
4 years ago
How many ions of aluminum oxide (Al²O³) are there in 200 g of Al²O³??<br><br>​
valentinak56 [21]

= 6.022 × 1020

Explanation<em>;</em>

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

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Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.051 molecules)

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020

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<em>hope </em><em>it </em><em>helps</em><em>_</em>

6 0
2 years ago
The boiling point of an aqueous solution is 101.02 °C. What is the freezing point?
kompoz [17]
Colligative properties calculations are used for this type of problem. Calculations are as follows:

ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
<span>ΔT(boiling point)  = (Kb)m
</span>m =  1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg

</span><span>ΔT(freezing point)  = (Kf)m
</span>ΔT(freezing point)  = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point)  = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
3 0
4 years ago
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