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denpristay [2]
3 years ago
13

How can scientists determine what kind of atoms are in stars and nebulas?

Chemistry
1 answer:
statuscvo [17]3 years ago
4 0
Each element absorbs light at specific wavelengths unique to that atom. When astronomers look at an object's spectrum, they can determine its composition based on these wavelengths.
Not sure if this is 100% but I hope it helps.
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What is the solubility of methylacetylene (in units of grams per liter) in water at 25 °C, when the C3H4 gas over the solution h
Luda [366]

Answer:

The solubility of methylacetylene is 0,11 g L⁻¹

Explanation:

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

The formula is:

C = kH P

Where C is solubility of the gas (In mol/L)

kH is Henry constant (9,23x10⁻² mol L⁻¹ atm⁻¹)

An P is partial pressure (0,301 atm)

Solving, C = 2,78x10⁻³ mol L⁻¹. In grams per liter:

2,78x10⁻³ mol L⁻¹ₓ \frac{40 g}{mol} = <em>0,11 g L⁻¹</em>

<em></em>

I hope it helps!

7 0
3 years ago
An element A has an atomic number of 11 and another element B has an atomic number 17 (A
MArishka [77]

Answer:

can anyone tell me how to get +7!67:$'!$

3 0
3 years ago
What happens with the energy radiated from the Sun once it gets to Earth? (mark ALL correct answers)
FrozenT [24]

Answer:

Absorbed by the surface of earth

Explanation:

Mark brainliest

3 0
2 years ago
How many grams of helium must be released to reduce the pressure to 65 atmatm assuming ideal gas behavior? Express the answer in
lyudmila [28]

The question is incomplete, here is the complete question:

How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.

<u>Answer:</u> The mass of helium released is 1.6 grams

<u>Explanation:</u>

We are given:

Mass of helium in the cylinder = 5.209 g

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{w}{M}RT

where,

P = Pressure of the gas  = 65 atm

V = Volume of the gas  = 334 mL = 0.334 L   (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = ?

M = Molar mass of helium gas  = 4 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gas = 23^oC=[23+273]K=296K

Putting values in above equation, we get:

65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g

Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g

Hence, the mass of helium released is 1.6 grams

5 0
3 years ago
What is a factor that affects the outcome of an experiment?
Anni [7]

Answer:

Observation affects the outcome

7 0
3 years ago
Read 2 more answers
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