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Lubov Fominskaja [6]
3 years ago
11

What cannot be changed by turning 100 grams of ice into water?

Chemistry
1 answer:
Yuliya22 [10]3 years ago
7 0
When you melt ice into water you should have less mass so it cant be a, the density is changed because the ice (a solid) turns into water (a liquid).  I think it should be phase because the volume should change when you melt ice,
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What is the key principle behind VSEPR theory
PSYCHO15rus [73]

Answer:

Minimizing repulsion by maximizing the bond angles

Explanation:

VSEPR is defined as 'valence shell electron pair repulsion' theory. The key principle is that repulsion of adjacent lone pairs of electrons or the electrons that are shared within a bond creates bond angles.

The bond angles are maximized in order to minimize the repulsion in the most stable geometrical shape of the molecule.

That said, in VSEPR theory we tend to achieve the greatest possible angles between the bonds and lone pair electrons in order to obtain the lowest repulsion between them.

8 0
3 years ago
A particular brand of gasoline has a density of 0.737 g/ml at 25 ∘c. how many grams of this gasoline would fill a 13.0 gal tank
Usimov [2.4K]

Density of gasoline is 0.737 g/mL and volume of tank is 13.0 gal.

Since, 1 US gal=3.78 L

Volume of tank in L will be:

V=13.0 gal(\frac{3.78 L}{1 gal})=49.14 L

Also, 1 L=1000 mL

Thus,

V=49.14 L(\frac{1000 mL}{1 L})=49140 mL

Mass of gasoline can be calculated as follows:

m=d×V

Here, d is density and V is volume thus,

m=0.737 g/mL\times 49140 mL=3.62\times 10^{4}g

Therefore, mass of gasoline will be 3.62\times 10^{4}g.

7 0
3 years ago
When a sulfur (s) atom becomes an ion, what charge does it usually have?
algol [13]

Then it would become: S2-

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6 0
3 years ago
Wally fluoride is an imaginary gaseous
mihalych1998 [28]

Answer:

\rho =1.96\frac{g}{L}

Explanation:

Hello there!

In this case, since this imaginary gas can be modelled as an ideal gas, we can write:

PV=nRT

Which can be written in terms of density and molar mass as shown below:

\frac{P}{RT} =\frac{n}{V} \\\\\frac{P}{RT} =\frac{m}{MM*V}\\\\\frac{P*MM}{RT} =\frac{m}{V}=\rho

Thus, by computing the pressure in atmospheres, the resulting density would be:

\rho = \frac{165/760 atm * 314.2 g/mol}{0.08206\frac{atm*L}{mol*K}*425K} \\\\\rho =1.96\frac{g}{L}

Best regards!

7 0
3 years ago
What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid require
laiz [17]

Explanation:

Below is an attachment containing the solution

4 0
3 years ago
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