What is the element with 2 protons on the

The central atom in ________ does not violate the octet rule. the central atom in ________ does not violate the octet rule. sf4

ruslelena [56]

We have to know which do not violet octet rule.

The central atom in SF₄ does not violate the octet rule. the central atom in CF₄ does not violate the octet rule.

As per octet rule, atoms can combine either by transfer of valence electrons from one atom to another or by sharing of valence electrons in order to have an octet in their valence shell.

The central atom S of SF₄ contains 8 electrons in the valence shell. Also, C atom of CF₄ contains 8 electrons in the valence shell.

ICl₄⁻ contains 9 electrons in the outermost shell.

7 0

3 years ago

Help this is due soon!!!

denis23 [38]

Answer:

not 100% sure but I think it's unbalanced chemical

6 0

4 years ago

Which linkages in starch and triglycerides result from condensation reactions?

Whitepunk [10]

Answer: The linkages in starch and triglycerides resulting from condensation reactions is glycosidic bonds in starch and ester bonds in triglycerides

8 0

4 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago

4. Nitrogen and oxygen gases react to form dinitrogen monoxide gas (N2O). What volume of O, is

solmaris [256]

Answer:

no be t ha t s had ice

Explanation:

uah an

8 0

3 years ago