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ValentinkaMS [17]
3 years ago
5

Why do electrons keep moving around the nucleus and not away from the atom

Chemistry
1 answer:
igomit [66]3 years ago
7 0
The atom's center, or nucleus, is positively charged and the electrons that whirl around this nucleus are negatively charged, so they attract each other. The reason the force is strong is because the atom is so small. The distance between the nucleus and the electrons is about 1 Angstrom (named after a famous scientist); this is 0.00000001 cm (10-8 cm) or about 4 billionths of an
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You have a 15.0 gram sample of gold at 20.0°C. How much heat does it take to raise the temperature to 100.0°C?
Nadusha1986 [10]

Answer:

=154.8 J

Explanation:

The rise in temperature is contributed by the change in temperature.

Change in enthalpy = MC∅,  where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.

Change in temperature = 100.0°C-20.0°C=80°C

ΔH=MC∅

The specific heat capacity of gold= 0.129 J/g°C

ΔH= 15.0g×0.129J/g°C×80°C

=154.8 J

7 0
3 years ago
The temparature of boiling water cannot be measured by _______ temparature
katrin [286]

Answer:

Temperature of boiling water cannot be measured by a Clinical thermometer The reason behind this is that the range of clinical thermometer varies between only 35° C to 42° C.

4 0
3 years ago
Read 2 more answers
Choose all that apply.
Mashutka [201]
A and D would be correct
5 0
3 years ago
Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa
Rashid [163]

Answer:

The temperature is 30,92K

Explanation:

We use the formula PV=nRT. We convert the unit of pressure in kPa into atm.

101,325kPa----1atm

121kPa-------x=(121,3kPax 1 atm)/101,325kPa=1, 2 atm

PV=nRT---->T= (PV)/(RT)

T=(1,2 atm x 3L)/(1,42 mol x 0,082 l atm/K mol )= 30, 91721058 K

6 0
3 years ago
Find the potentials of the following electrochemical cell:
melomori [17]

Answer: 0.18 V

Explanation:-

Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0_(Cd^{2+}/Cd)=-0.40V[/tex]

E^0_(Ni^{2+}/Ni)=-0.24V[/tex]

Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}

E^0=-0.24-(-0.40)=0.16V

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential = 0.16 V

E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}

E_{cell}=0.18V

Thus the potential of the following electrochemical cell is 0.18 V.

6 0
3 years ago
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