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3 years ago

what will happen to the nucleus if the attractive forces in the nucleus are weaker than repulsive forces?

Chemistry

1 answer:

RideAnS [48]3 years ago

7 0

Everything would implode

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How many atoms are in 4H3BO3

maksim [4K]

1 mole of h3bo3...........6.023*10²³ each h and B and 0 so we will have 3hydrogen+ 1 B+3 oxygen = 7*6.023*10²³ atoms

1 mole .......7*6.023*10²³atoms

4 moles ........x atoms

x=4*7*6.023*10²³.

3 0

3 years ago

Which of the following shows the valency of 1?

Oliga [24]

$\huge \bf༆ Answer ༄$

The element having valency of 1 is ~

Sodium (Na)

6 0

3 years ago

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The equilibrium expression for a reaction is `"K"_("eq") = ("[H"^+"]"^6)/("[Bi"^(2+)"]"^2["H"_2"S"]^3)` Which of the following c

Nata [24]

Option B is correct

K = Kp /Kr

The given equation indicating, the product containing 6 moles of proton whereas the reactant contains 2 mole of bismuth and 3 mole of hydrogen sulphide.

Hence, in reaction B there are 2 mole of bismuth and 3 mole of hydrogen sulphide reacting to produce 6 moles of proton. whereas the concentration of Bi2S3 is not considered as it is present in solid phase.

4 0

3 years ago

Read 2 more answers

Can you explain what he did wrong.

vivado [14]

Cant see it well please find better view

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