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1 year ago

PLEASE ANSWER AS QUICKLY AS POSSIBLE THANK YOU SO MUCH

Chemistry

1 answer:

Harlamova29_29 [7]1 year ago

6 0

Answer:

This description needs a unit describing the system of measurement.

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The maximum number of electrons in a single d subshell is:

Iteru [2.4K]

10 electrons

Explanation:

The maximum number of electrons in a single d-subshell is 10 electrons.

The d-sub-orbital used to denote azimuthal or secondary quantum numbers.

The maximum number of electrons in the orbitals of sublevels are:

two electrons in the s-sublevel, it has one orbital

six electrons in the p-sublevel, it has three orbital

ten electrons in the d- sublevel, it has five orbitals

fourteen electrons in the f-sublevel, it has seven orbitals

The maximum number of electrons in an orbital is two.

learn more:

Atomic orbitals brainly.com/question/1832385

#learnwithBrainly

4 0

3 years ago

Gimme ur bath water plzzzz

kari74 [83]

Answer:

plzz can you explain in detail this Q

6 0

3 years ago

How do you solve this ?NH4+NO4->N2+O2+H2O2

ololo11 [35]

I'm assuming you wanted this equation balanced? If you typed it correctly, it's already balanced for you. :)

8 0

3 years ago

Can anyone please help with this!!

Sphinxa [80]

Answer:

2341, last option is the correct choice.

Explanation:

Boiling points of the given compounds are given as:

$CH_4=-161.5\:C^{\circ}\\\\C_4H_{10}=-1\:C^{\circ}\\\\CH_2Cl_2=39.6\:C^{\circ}\\\\H_2O=100\:C^{\circ}$

Best Regards!

5 0

3 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

Now put all the given values in the above formula, we get:

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0

3 years ago