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Alborosie
3 years ago
5

Which of the following is an example of a contact force?

Chemistry
2 answers:
Virty [35]3 years ago
7 0
C. A person pushing a box across the floor
MAXImum [283]3 years ago
5 0
I believe it would be C. A person pushing a box across the floor because it is the only thing that uses direct contact to apply a force, which is the definition of a contact force
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What type of bonding occurs in a sample of pure chromium, cr? in other words, how is one chromium atom held to another chromium
spayn [35]

Chromium is a metal in nature. So when one chromium is bonded to another chromium, there is a weak intermolecular forces which helds them together which we call as “metallic bonding”.

Metallic bonding is the intermolecular force of attraction which exist between valence electrons and the metal atoms. It is considered as the sharing of various detached electrons between many positive ions, whereby the electrons serve as a "glue" which gives the substance a definite structure.

4 0
3 years ago
Chemistry screen shot below plzzzzzzz help i've been stuck forever
Oxana [17]

Answer:

True

Explanation: Imagine the Electrons is by the nucleus which give more energy.

8 0
3 years ago
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What are the metalloids on the periodic table?
Sauron [17]
<span>The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At) are the elements found along the step like line between metals and non-metals of the periodic table.</span>
5 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
4 years ago
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What is the ideal gas law
dlinn [17]

Answer: Gases are complicated. They're full of billions and billions of energetic gas molecules that can collide and possibly interact with each other. Since it's hard to exactly describe a real gas, people created the concept of an Ideal gas as an approximation that helps us model and predict the behavior of real gases. The term ideal gas refers to a hypothetical gas composed of molecules which follow a few rules:

Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container. [What is an elastic collision?]

Ideal gas molecules themselves take up no volume. The gas takes up volume since the molecules expand into a large region of space, but the Ideal gas molecules are approximated as point particles that have no volume in and of themselves.

If this sounds too ideal to be true, you're right. There are no gases that are exactly ideal, but there are plenty of gases that are close enough that the concept of an ideal gas is an extremely useful approximation for many situations. In fact, for temperatures near room temperature and pressures near atmospheric pressure, many of the gases we care about are very nearly ideal.

If the pressure of the gas is too large (e.g. hundreds of times larger than atmospheric pressure), or the temperature is too low (e.g.

−

200

C

−200 Cminus, 200, start text, space, C, end text) there can be significant deviations from the ideal gas law.

Explanation:

7 0
3 years ago
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