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padilas [110]
3 years ago
13

According to the bohr model of the atom, where can atoms exist? where will they ot exist

Chemistry
1 answer:
Inga [223]3 years ago
6 0
At least, that's what Bohr<span> decided, and that's why he proposed the </span>existence<span> of the</span>atomic<span> energy level. </span>According<span> to </span>Bohr<span>, the electrons in an </span>atom<span> were only allowed to </span>exist<span> at certain energy levels</span>
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Help out please?????
Alexxandr [17]

Answer:

A non- metal

Explanation:

Hope u got ur answer

5 0
3 years ago
Read 2 more answers
What is the mass of calcium in 2.3•10^23 molecules of calcium phosphate
Katena32 [7]

Explanation:

N(Ca)=3×N(Ca₃(PO₄)₂)=

=3×2.3×10²³=6.9×10²³

6.02×10²³→40g

6.9×10²³→Xg

m(Ca)=X≈45.85g

3 0
3 years ago
An average bushel of apples has a mass of 10 kg and contains 5 dozen apples.
Alina [70]

Answer:

Average mass of 6 apple  = 1 kg

Explanation:

Given:

Average mass of 5 dozen apple = 10 kg

Find:

Average mass of 6 apple

Computation:

Average mass of 6 apple  = [Average mass of 1 apple] x [6 apple]

Average mass of 6 apple  = [10/(12 x 5)] x [6]

Average mass of 6 apple  = 1 kg

3 0
3 years ago
Calculate the volume of the gas, in liters, if 1.75 mol has a pressure of 1.28 atm at a temperature of -7 ∘C
Alexandra [31]

Answer:

A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.

 

You need to apply the ideal gas law PV=nRT

 

You have the pressure, P=1.01 atm

you have the volume, V = 2.21 L

The ideal gas constant R= 0.08205 L. atm/ mole.K at  273 K

 

find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)

 

n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and

L=1.23 L

 

P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L

= 1.994 atm

Explanation:

6 0
3 years ago
How many moles are in 15.0 grams of calcium (Ca)?
borishaifa [10]
1 mol ---- 40g
X --------- 15g
X = 15/40 = 0,375 moles

:)
5 0
3 years ago
Read 2 more answers
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