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4 years ago

7

Which contributes to the dissolution of sugar in water?

2 answers:

Feliz [49]4 years ago

7 0

D would be my best bet.

erma4kov [3.2K]4 years ago

4 0

The sugar is a polar molecule and same as the water. The polar water molecule will attract the positive and negative part of the sugar and lead the sugar dissolve.

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SOMEONE HELP ILL BRAINLISTTTTTTY. Write in your own wordsss. I’ll give you 70 points and a brainlist in total. Which is enough w

ZanzabumX [31]

Answer:

Explanation:

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3 years ago

Atomic Number: 10<br> Is:<br> 2s:<br> 2p:<br> 35:<br> 3p:<br> 4s:<br> 3d:<br> 4p:<br> 5s:

aivan3 [116]

Answer:

Atomic Number: 10

Is: Neon

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3 years ago

Which of these metals reacts more vigorously with water? <br><br> 1. Zinc <br> 2. Magnesium

GalinKa [24]

Answer:

magnesium

Explanation:

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3 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

<u>Answer:</u> The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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3 years ago

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MAVERICK [17]

C) can give up electrons more readily than others.

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3 years ago

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