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lesya692 [45]
3 years ago
11

Which incorrect aspect of Rutherford’s model was fixed by Bohr’s model?

Chemistry
1 answer:
dolphi86 [110]3 years ago
7 0

Answer:

Some things that were wrong with Rutherford's model were that the orbiting electrons should give off energy and eventually spiral down into the nucleus, making the atom collapse. Bohr proposed his quantized shell model of the atom to explain how electrons can have stable orbits around the nucleus. To remedy the stability problem, Bohr modified the Rutherford model by requiring that the electrons move in orbits of fixed size and energy.

Explanation:

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Matter must have two physical properties: 1. have mass, and 2.
djverab [1.8K]
I believe the answer is density

6 0
3 years ago
A 111.6 gram sample of iron (MW=55.8) was heated from 0 degrees C to 20 degrees C. It absorbed 1004 Joules of energy. What is th
qwelly [4]
Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity 
ΔT=change in temperature (in this case 20°C)
You have to rewrite the equation for c.
c=q/nΔT
c=1004J/(2.08mol x 20°C)
c=24.1 J/mol°C
 
I hope this helps
7 0
3 years ago
What is the mass in grams of 6.12 x 10^20 molecules of caffeine, C8H10N4O2?
baherus [9]
6,02*10^{23} \ \ \ \ \ \ \rightarrow \ \ \ \ \ 194,2g\\
6,12 * 10^{20} \ \ \ \ \ \ \rightarrow \ \ \ \ \ x\\\\
x=\frac{6,12*10^{20}*194,2g}{6,02*10^{23}}\approx197*10^{-3}g=0,197g
6 0
3 years ago
A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
What is Potassium Oxalate name formula?
melisa1 [442]

Answer:

formula pottasium Oxalate C2k204

Explanation:

C2k204

6 0
3 years ago
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