I ASKED THIS QUESTION OVER 20 TIMES PLEASE HELP

Help me with this chemistry question please.

rodikova [14]

Answer:

B) Because you are adding positively charged protons, which pull in the outer electrons.

Explanation:

Since the increase in protons have a greater effect then the increase in electrons the protons pull the electrons in closer resulting the radius of an atom to decrease from left to right across a period.

4 0

4 years ago

When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu

Gwar [14]

Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

Moles S₈: 256.52g/mol

10.0g * (1mol / 256.52g) = 0.0390 moles

Moles F₂:

PV = nRT

PV/RT = n

Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%

6 0

3 years ago

The carbon bond strengths of graphite and diamonds are identical. True or False

Elena-2011 [213]

Answer:

false :)

Explanation:

3 0

4 years ago

Read 2 more answers

How many significant figures are in 675000 m 6 4 3 5

Dima020 [189]

Answer:

There are 3 significant figures in 675000

Explanation:

The rules for significant figures:

- Non zero digit (from 1 upward) are always significant

- Trailing zero are not significant

- but zero in the presence of decimal are significant

In this case, the trailing zeros are not significant thus only the first 3 non zero digit are significant.

4 0

3 years ago

Give the formulas of the compounds in each set (a) lead(l|) oxide and lead(lV) oxide; (b) lithium nitride, lithium nitrite, and

Ivanshal [37]

Answer:

For a: The formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b: The formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c: The formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d: The formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

Explanation:

All the given compounds are ionic compounds. This means that between the atoms complete sharing of electrons takes place.

For a:

Lead is the 82th element of periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^2$ .

To form $Pb^{2+}$ ion, this element will loose 2 electrons and to form $Pb^{4+}$ ion, this element will loose 4 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lead (II) oxide is PbO and for lead (IV) oxide is $PbO_2$

Thus, the formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b:

Lithium is the 3rd element of periodic table having electronic configuration of $[He]2s^1$ .

To form $Li^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

Nitrite ion is a polyatomic ion having chemical formula of $NO_2^{-}$

Nitrate ion is a polyatomic ion having chemical formula of $NO_3^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lithium nitride is $Li_3N$ , for lithium nitrite is $LiNO_2$ and for lithium nitrate is $LiNO_3$

Thus, the formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c:

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Hydrogen is the 1st element of periodic table having electronic configuration of $1s^1$ .

To form $H^{-}$ ion, this element will gain 1 electron and is named as hydride ion.

Hydroxide ion is a polyatomic ion having chemical formula of $OH^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium hydride is $SrH_2$ and for strontium hydroxide is $Sr(OH)_2$

Thus, the formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d:

Magnesium is the 12th element of periodic table having electronic configuration of $[Ne]3s^2$ .

To form $Mg^{2+}$ ion, this element will loose 2 electrons.

Manganese is the 25th element of the periodic table having electronic configuration of $[Ar]3d^54s^2$

To form $Mn^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium oxide is MgO and for manganese (II) oxide is MnO.

Thus, the formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

7 0

3 years ago