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Taya2010 [7]
3 years ago
7

Reactivity, flammability, and the ability to rust are types of _________.

Chemistry
2 answers:
vaieri [72.5K]3 years ago
8 0

Answer:

stink

Explanation:

ra1l [238]3 years ago
8 0
Physical changes is the correct answer I believe
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Dhhdhd d jejnen r nejr
Marianna [84]

Answer:

Okay so my wifi is slow. How are you!

Explanation:

I don't see an image so I assume it's free points.....thank you! But if there is an image let me know! and Brainliest?? Pretty please!

8 0
3 years ago
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Is there a way to dissolve more sugar into a solution that it should hold at that temperature? Explain how this is done and what
Darya [45]

The answer is: supersaturated solution.

A supersaturated solution contains more of the dissolved substance than could be dissolved by the solvent under normal circumstances.

A way to dissolve more sugar into a solution is heating a solution.

The more heat is added to a system, the more soluble a substance (in this example sugar) becomes.

The solution will become supersaturated if this solution is suddenly cooled at a rate faster than the rate of precipitation.

6 0
3 years ago
The pictures to the right, show two different models of the atom. Type in the letter of your answer. Which model best represents
Solnce55 [7]

Answer:

Model B

Explanation:

Just did it on Edgenuity.

6 0
3 years ago
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When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
What are other portals of entry besides needlesticks?
aivan3 [116]

Answer:

Inhalation (via the respiratory tract)Absorption (via mucous membranes such as the eyes) Ingestion (via the gastrointestinal tract)

Explanation:

The opening where an infectious disease enters the host's body such as mucus membranes, open wounds, or tubes inserted in body cavities like urinary catheters or feeding tubes.

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