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3 years ago

Name three stars that are hotter than the sun

1 answer:

7 0

Eta Carinae could be as large as 180 times the radius of the Sun, and its surface temperature is 36,000-40,000 Kelvin. Just for comparison, 40,000 Kelvin is about 72,000 degrees F. So it's the blue hypergiants, like Eta Carinae, which are probably the hottest stars in the Universe.

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What phases are most common at high temperatures

Nikitich [7]

Answer:

Liquids and gases.

For most substances, when heated to a high enough temperature, the molecules begin moving faster which results in its liquid or gas state.

5 0

2 years ago

Please help asap :)))))

deff fn [24]

Answer:

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6 0

2 years ago

The number of atoms or molecules in a mole is known as the __________ constant.

AnnZ [28]

Answer: Avogrado's Constant

Explanation:

One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles.

7 0

2 years ago

Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0

2 years ago

A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect

rosijanka [135]

Answer:

38.3958 °C

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories of energy

1 gram of protein on burning gives 4 kilocalories of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy

Total energy = 515 kilocalories

Using,

$Q=m_{water}\times C_{water}\times (T_f-T_i)$

Given: Volume of water = 23 L = 23×10⁻³ m³

$Density=\frac{Mass}{Volume}$

Density of water= 1000 kg/m³

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}$

Mass of water = 23 kg

Initial temperature = 16°C

Specific heat of water = 0.9998 kcal/kg°C

$515=23\times 0.9998\times (T_f-16)$

Solving for final temperature as:

<u>Final temperature = 38.3958 °C </u>

8 0

3 years ago