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Hitman42 [59]
3 years ago
7

6. What happens to the arrangement of water molecules as ice melts?

Chemistry
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

D. Molecules are able to move as they heat up to melt

Explanation:

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Identify the covalent compounds based on the names of the compounds. barium nitrate dinitrogen tetroxide boron trifluoride ammon
noname [10]

Covalent compounds are composed of atoms that are linked via covalent bonds i.e. bonds formed by mutual sharing of electrons. This is in complete contrast to ionic compounds which are held together by ionic bonds, i.e. bonds formed by complete transfer of electrons from one atom to the other.

In the given examples we have:

Barium nitrate: Ba(NO3)2 - Ionic

Dinitrogen tetroxide: N2O4- Covalent

Boron trifluoride: BF3-Covalent

Ammonium sulfate: (NH4)2SO4- Ionic

Carbon tetrachloride: CCl4- Covalent

Barium chloride: BaCl2 - Ionic



5 0
3 years ago
Read 2 more answers
Help me with this question please.
Salsk061 [2.6K]

Explanation:

current =  \frac{voltage}{resistance}  \\ i = \frac{v}{ohm}  \\  i  = \frac{240}{38}  \\ i = 6.3158 \: amps

hope it helped

pls mark BRAINLIEST

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8 0
3 years ago
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
What is the mass, in grams, of 28.58 mL of acetone?
Vedmedyk [2.9K]
<span>If you look up the density of Acetone (Propanone in IUPAC names) you will find it is 0.7925g/cm3. This is the same as 0.7925g/ml. 

You can calculate mass using the equation:- mass = density x volume 
In your example mass = 0.7925 x 28.40 = 22.51g</span><span>
I think That's right. Hope this helps!!! Good luck!</span>
7 0
3 years ago
Read 2 more answers
HURRY HELP 10 POINTS ILL GOVE BRAINLIEST I HAVE TO PASS PICTURE PROVIDED
kramer

Answer:

C₃H₄O₄

Explanation:

In order to get the empirical formula of a compound, we have to follow a series of steps.

Step 1: Divide the percent by mass of each element by its atomic mass.

C: 34.6/12.01 = 2.88

H: 3.9/1.01 = 3.86

O: 61.5/16.00 = 3.84

Step 2: Divide all the numbers by the smallest one, i.e., 2.88

C: 2.88/2.88 = 1

H: 3.86/2.88 ≈ 1.34

O: 3.84/2.88 ≈ 1.33

Step 3: Multiply all the numbers by a number that makes all of them integer

C: 1 × 3 = 3

H: 1.34 × 3 = 4

O: 1.33 × 3 = 4

The empirical formula is C₃H₄O₄.

5 0
3 years ago
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