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Harman [31]
3 years ago
10

Which substance is an example of a nonpolar compound?(1 point)

Chemistry
1 answer:
Tatiana [17]3 years ago
8 0
The answer is B - vegetable oil.

Explanation:

The properties of liquids depend on the attractions the molecules of the liquid have for each other and for other substances.

Liquids can dissolve certain other liquids, depending on the attractions between the molecules of both liquids.

Polar liquids, like water, dissolve other liquids which are polar or somewhat polar.

Polar liquids, like water, do not dissolve non-polar liquids like oil.
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2. Which test for iron(II) ions is conclusive ​
krok68 [10]

Answer:

please brainlist answer

Explanation:

The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.

5 0
3 years ago
Read 2 more answers
how do the boiling point and freezing point of a solution of water and calcium chloride at standard pressure compare to the boil
Snezhnost [94]
Both the increase in the boling point and the depression on the freezing point are colliative properties.

This is, they are proportional to the number of particles dissolved in the solvent, which is measured by the molality of the solution and the factor i (Van'f Hoff).

The answer to the question is that 1) the boling point of a solution of water and calcium chloride at standard pressure will be higher than the normal boiling point of pure water, and 2) the freezing point of a solution of water and calcium chloride at standard pressure will be lower than the normal freezing point of pure water.
3 0
3 years ago
What is the shape of graphite​
VashaNatasha [74]

Answer:

The molecular formula lists the symbol of each element within the compound followed by a number (usually in subscript). The letter and number indicate how many of each type of element are in the compound. If there is only one atom of a particular element, then no number is written after the element.

7 0
3 years ago
SOMEONE PLEASE HELP
Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

= 8.16 \times 10^{ - 6} \:  kg

\boxed{density =  \frac{mass}{volume} }

Density of iron bar

=  \frac{0.0642}{8.16 \times 10^{ - 6} }

= 7870 kg/m³ (3 s.f.)

7.

\boxed{mass = density \:  \times volume}

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

= 12  \times  10^{ - 6}  \: m^{3}  \\  = 1.2 \times 10^{ - 5}  \: m ^{3}

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

=  \frac{0.151} {1.25 \times 10^{ - 5} }  \\  = 12600 \: kg/ {m}^{3}

(3 s.f.)

6 0
3 years ago
What is #6 ,7,8 need help
weeeeeb [17]
#6 should be the independent variable because that's the one you can control
6 0
3 years ago
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