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Nookie1986 [14]
4 years ago
5

What forces pushes protons in the nucleus apart

Chemistry
2 answers:
Nimfa-mama [501]4 years ago
7 0
Bring that ahs here boy 
alisha [4.7K]4 years ago
6 0
<span>Nuclear </span>force<span> is what holds atomic </span>nuclei<span> together in spite of the repulsive electromagnetic </span>force <span>between the positively charged protons that acts to </span>push<span> them </span>apart<span>.</span>
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Which has a longer wave length visable or infrared
lutik1710 [3]

Answer:

Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. Infrared waves have wavelengths longer than visible and shorter than microwaves, and have frequencies which are lower than visible and higher than microwaves

Explanation:

4 0
3 years ago
Generally, when going down a group on the periodic table:
antiseptic1488 [7]

The correct answer is ionic radii increase.  

The ionic radii decrease as one move across the periodic table, that is, from left to right, while the ionic radius increases as one move from top to bottom on the periodic table. As one moves down a group in the periodic table, the supplementary layers of electrons are being added that usually results in the increase of the ionic radius as one moves down the periodic table.  


7 0
3 years ago
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
4 years ago
An ideal monatomic gas at 300 K expands adiabatically to twice its initial volume. Assume that the process is also quasistatic.
kondor19780726 [428]

Answer:

600k

Explanation:

In this problem, we need to use a gas law that relates temperature to volume. The gas law to use here is the Charles’ law.

The Charles’ law posits that temperature and volume are directly proportional, provided that the pressure is kept constant.

Mathematically:

V1/T1 = V2/T2

We are looking at getting V2, hence we can write the mathematical equation as:

T2 = V2T1/V1

Asides the fact that we know that the gas is monoatomic, we do not know its volume. Let its initial volume be v. Since it expanded adiabatically, this means that its new volume is 2v

Hence: V1 = v , V2 = 2v , T1 = 300k and T2 is ?

Substituting these values, we have the following:

T2 = (2v * 300)/v

T2 = 600k

6 0
4 years ago
For each molecule below draw the structural formula of the molecule
torisob [31]
I think you forgot to post a picture
5 0
4 years ago
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