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Maslowich
3 years ago
14

Help out please?????

Chemistry
2 answers:
strojnjashka [21]3 years ago
7 0
A Non-Metal obviously
Alexxandr [17]3 years ago
5 0

Answer:

A non- metal

Explanation:

Hope u got ur answer

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PLEASE HELP GUYS PLEASE!!!! I will LOVE WHOEVER HELP<br> ME FOREVER
Vladimir79 [104]

Answer:

B. It is highly flammable.

Explanation:

Chemical properties deal with how the substance will behave when brought into contact with other substances, and usually have to do with it undergoing some kind of reaction. Flammability is a chemical property.

Density, melting point, and color are all considered physical properties.

3 0
3 years ago
Read 2 more answers
A student conducts an experiment with a steel wool pad, which is comprised of iron. The student washes the steel wool in vinegar
Ksju [112]
B. A chemical change has occurred and energy was given off
3 0
3 years ago
Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘C. Express the entropy change to
zhuklara [117]

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

8 0
3 years ago
Substances that are readily combustible or may cause fire through friction
Dvinal [7]

Answer :

Flammable substances

Explanation :

<em>Flammable substances</em> will catch fire and continue to burn when they contact an ignition source like a spark or a flame.

For example, <em>methanol</em> is a flammable liquid.

A flammable solid may also catch fire through friction. <em>Matches</em> are flammable solids.

3 0
3 years ago
Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of
pantera1 [17]

Answer:

W=5743.1077\ J

Explanation:

The expression for the work done is:

W=RT \ln \left( \dfrac{P_1}{P_2} \right)

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

T is the temperature

P₁ is the initial pressure

P₂ is the final pressure

Given that:

T = 300 K

P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)

W=300\times \:8.314\ln \left(10\right)

W=2.30258\times \:2494.2

W=5743.1077\ J

5 0
3 years ago
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