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slega [8]
3 years ago
7

HELP!!! Question 10.

Chemistry
2 answers:
umka2103 [35]3 years ago
6 0
Stationary front is the answer
salantis [7]3 years ago
5 0
It’s stationary front.
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If you can smell your grandmother’s perfume from across the room, you are experiencing the result of which liquid property?
cestrela7 [59]
The appropriate answer is D. volatility. Volatility refers to the susceptibility of liquids to vaporize. Perfume is liquid when applied but because of volatility, it has a tendency to vaporize and so it will convert to a gas and diffuse across the room. The process by which a liquid changes to a vapor is called evaporation. 
5 0
2 years ago
We gave an object with a density of 620 g/cm^3 and a volume of 75cm^3. What is the mass of this object?
Hitman42 [59]
d=\frac{m}{V} \ \ \ \Rightarrow \ \ m=dV\\\\
d=620\frac{g}{cm^{3}}\\
V=75cm^{3}\\\\\
m=620\frac{g}{cm^{3}}*75cm^{3}=46500g
6 0
3 years ago
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 0
3 years ago
At 333 k, which of the pairs of gases below would have the most nearly identical rates of effusion?
rjkz [21]
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
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b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
8 0
3 years ago
How can these objects all weigh the same but be different sizes?
Mrrafil [7]
Density/Earth’s gravitational pull.
5 0
2 years ago
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