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Morgarella [4.7K]
3 years ago
12

Help plsss

Chemistry
2 answers:
Tema [17]3 years ago
5 0
#47 is New Mexico, and its capital is Santa Fe.

Happy studying ^-^
PtichkaEL [24]3 years ago
3 0
I think number 47 on the map would be New Mexico i hope i helped out
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A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

7 0
3 years ago
How many molecules are contained in 125 grams of water, H20?
sergejj [24]
Answer:

18,01528

Explanation:
8 0
3 years ago
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An atom with 13 protons and an unknown number of neutrons is called____?
MaRussiya [10]
Aluminum contains 13 protons
4 0
3 years ago
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A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius
Nonamiya [84]
Data Given:
                  Pressure  =  P  =  ?

                  Volume  =  V  =  3.0 L

                  Temperature  =  T  =  115 °C + 273  =  388 K

                  Mass  =  m  =  75.0 g

                  M.mass  =  M  =  44 g/mol

Solution:
              Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
                                      P V  =  n R T
Solving for P,
                                      P  =  n R T / V      ------ (1)
Calculating Moles,
                                      n  =  m / M

                                      n  =  75.0 g / 44 g.mol⁻¹

                                      n  =  1.704 mol

Putting Values in Eq. 1,

                    P  =  (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

                    P  =  18.08 atm
7 0
3 years ago
Which of the following would you except to see in the death of a star that is less than 0.5 solar mass
Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: 1.4 \;M_\odot.

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 0.5 \;M_\odot, which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0
3 years ago
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