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horsena [70]
2 years ago
10

B) The size of a star is a balance between 2 things. Explain this statement.

Chemistry
1 answer:
Zielflug [23.3K]2 years ago
8 0

Answer:

Luminosity is the amount of light that a star radiates. The size of the star and its surface temperature determine its luminosity. Apparent magnitude of a star is its perceived brightness, factoring in size and distance, while absolute magnitude is its true brightness irrespective of its distance from earth

Explanation:

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If you want 1,775 mL of 48 M MgCl2 solution, how many grams of MgCl2 do you need?
vlada-n [284]

Answer:

0.1

Explanation:

3 0
2 years ago
What is the amount of aluminum chloride produced from 40 moles of chlorine and excess aluminum?
Vera_Pavlovna [14]

Answer: Molar mass of Al = 26.98 g/mol

mass of Al = 12 g

mol of Al = (mass)/(molar mass)

= 12/26.98

= 0.4448 mol

According to balanced equation

mol of AlCl3 formed = moles of Al

= 0.4448 mol

Answer: 0.445 mol

hope this help boo❤️❤️❤️

Explanation:

3 0
3 years ago
I need to make up an experiment !! Please help
katen-ka-za [31]

Question: Baking a Cake Without Flour.

Hypothesis: I think that when I remove the flour from the standard cake recipe, I'll end up with a flat but tasty cake.

Procedure: I baked two cakes during my experiment. For my control, I baked a cake following a normal recipe. I used the Double Fudge Cake recipe on page 292 of the Betty Crocker Cookbook. For my experimental cake, I followed the same recipe but left out the flour. I first obtained a 2-quart mixing bowl.  

Results: My control cake, which I cooked for 25 minutes, measured 4 cm high.  Eight out of ten tasters that I picked at random from the class found it to be an acceptable dessert. After 25 minutes of baking, my experimental cake was 1.5 cm high and all ten tasters refused to eat it because it was burnt to a crisp.

What did I learn?/Conclusion: Since the experimental cake burned, my results did not support my hypothesis.  I think that the cake burned because it had less mass, but cooked for the same amount of time.  I propose that the baking time be shortened in subsequent trials.

-

I hope this helped :))

7 0
3 years ago
If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit
Blizzard [7]

<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

<u>Explanation:</u>

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

V=10.00\times 7.25\times 3.00=217.5cm^3

  • <u>For a:</u>

To convert the volume of cuboid into milliliters, we use the conversion factor:

1mL=1cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL

Hence, the volume of the box is 217.5 mL

  • <u>For b:</u>

To convert the volume of cuboid into liters, we use the conversion factor:

1L=1000cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L

Hence, the volume of the box is 0.2175 L

3 0
3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
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