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Andre45 [30]
3 years ago
13

If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property

Physics
1 answer:
Ainat [17]3 years ago
3 0

Answer:

  • <u><em>soluble</em></u>

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a<em> physical property</em> because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is <em>soluble</em> is describing a <em>physical property.</em>

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Answer:

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Explanation:

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2 years ago
Earth’s atmosphere is in hydrostatic equilibrium. What this means is that the pressure at any point in the atmosphere must be hi
Misha Larkins [42]

Answer: The pressure that one experiences on the Mount Everest will be different from the one, in a classroom. It is because pressure and height are inversely proportional to each other. This means that as we move up, the height keeps on increasing but the pressure will keep on decreasing. This is the case that will be observed when one stands on the Mount Everest as the pressure is comparatively much lower there.

It is because as we move up, the amount of air molecules keeps on decreasing but all of the air molecules are concentrated on the lower part of the atmosphere or on the earth's surface.

Thus a person in a low altitude inside a classroom will experience high pressure and a person standing on the Mount Everest will experience low pressure.

6 0
3 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
A student performs a reaction twice. In the second trial, she raises the temperature by 20¨¬C and notices that the reaction take
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Endothermic reactions: The reactions in which the energy of the products is more than the energy of the reactants, and the excess energy is absorbed as heat.

3 0
3 years ago
Read 2 more answers
The input work done on a machine is 9.63 × 103 joules, and the output work is 3.0 × 103 joules. What is the percentage efficienc
notsponge [240]
Input work = 9.63×10³ J.
Output work = 3.0×10³ J

By definition,
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Answer:  31%
5 0
3 years ago
Read 2 more answers
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