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SVEN [57.7K]
3 years ago
5

Describe the structure of our galaxy, including its spiral arms.

Chemistry
2 answers:
gulaghasi [49]3 years ago
6 0

Answer:

Most spiral galaxies consist of a flat, rotating disk containing stars, gas and dust, and a central concentration of stars known as the bulge. ... The spiral arms are sites of ongoing star formation and are brighter than the surrounding disc because of the young, hot OB stars that inhabit them.

Explanation:

vlada-n [284]3 years ago
4 0

Answer:

Unlike a regular spiral, a barred spiral contains a bar across its center region, and has two major arms. The Milky Way also contains two significant minor arms, as well as two smaller spurs. One of the spurs, known as the Orion Arm, contains the sun and the solar system.

Explanation:

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4. What is the difference between an open system and a<br> closed system?
Nimfa-mama [501]

Systems can exist in three ways as open systems, closed systems, and isolated systems. The main difference between open and closed system is that in an open system, matter can be exchanged with the surrounding whereas, in a closed system, matter cannot be exchanged with the surrounding.

btw why not just look it up on google? btw btw brainliest plz

5 0
3 years ago
How many moles of lithium are in 18.2 grams of lithium
Doss [256]

Answer:

I think 2.6 sorry if its wrong

6 0
3 years ago
Read 2 more answers
For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

3 0
3 years ago
Read 2 more answers
At what the pressure, in atmospheres, that O2, gas at 75°C has a density of 2.00 g/L​
kari74 [83]
Pink and fluffy and squishy mangos
8 0
3 years ago
What else is produced during the combustion of butane, C4H10?
denpristay [2]

Another product: CO₂

<h3>Further explanation</h3>

Given

Reaction

2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O

Required

product compound

Solution

In the combustion of hydrocarbons there can be 2 kinds of products

If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product

If Oxygen is low, you'll get Carbon monoxide(CO) and water

Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products

if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left

2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O

5 0
2 years ago
Read 2 more answers
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