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mariarad [96]
2 years ago
7

Mixed Practice:

Chemistry
1 answer:
tigry1 [53]2 years ago
4 0

Given

Mass of NO - 824 g

Molar mass of NO - 30.01g/mol

No of moles of NO = Given mass/Molar mass

No of moles of NO = 824/30.01= 27.45 mole

Hence 27.5 moles of NO are formed!

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Which oneeee is ittt!!!!
Marysya12 [62]

Answer:

Disagree.

Explanation:

Nothing would remain. Everything on Earth is made up of atoms. The chair consists of only atoms. So, if you remove all of the atoms, nothing would be left! Hope this helps.

7 0
3 years ago
Read 2 more answers
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

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

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
Rotation about a carbon-carbon double bond does not readily occur because: __________.1) the overlap of the p orbitals of the ca
Advocard [28]

Answer:

1) The overlap of the p orbitals of the carbon-carbon π bond would be lost

Explanation:

Unlike simple bonds, a double bond can not rotate, since it is not possible to twist the ends of the molecule without breaking the π bond.

In the structure of but-2-ene present in the attachment, we can see the two isomers, <em>cis</em> and<em> trans</em>. These isomers cannot be interconverted by rotation around the carbon-carbon double bond without breaking the π bond.

5 0
3 years ago
How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *
serg [7]

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

6 0
2 years ago
How many mL of a 4% mass/volume Mg(NO3)2 solution would contain 1.2 grams of magnesium nitrate?
Phoenix [80]
4% mass / volume :

4 g ---------> 100 mL
1.2 g ------- ? mL

V = 1.2 * 100 / 4

V = 120 / 4

V = 30 mL

hope this helps!

7 0
3 years ago
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