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Harrizon [31]
2 years ago
15

Makes up the core of the other planets stars with an M and ends with E

Chemistry
2 answers:
Gwar [14]2 years ago
4 0

Answer: I think the M would be Mars But I am not sure what the L would be. Sorry.

Explanation: Hope that helps!

Sonbull [250]2 years ago
3 0

Answer:

its iron not m or e

Explanation:

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3 years ago
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What color would you expect a weak base to turn the paper?
Darina [25.2K]

Answer:

Blue litmus paper paper turns red under acidic conditions. Red litmus paper turns blue under basic or alkaline conditions. If you're using a universal indicator, the weak base should turn those blue.

Explanation:

7 0
3 years ago
Describe the stages of volcanic activity
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3 years ago
What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?
Brut [27]

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

7 0
3 years ago
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave
timofeeve [1]

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

n_i=8

\lambda=3745\ nm

The amount of energy change during the transition is given by :

\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]

And

\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]

Plugging all the values we get :

\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5

So, the final level of the electron is 5.

4 0
3 years ago
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