Which one would it be <br> 1,2,3,4

What heat is liberated in the formation of 10.0 grams of sulfur hexafluoride, SF6, from the elements sulfur and fluorine?

tensa zangetsu [6.8K]

Answer:

$Q=-76.7kJ$

Explanation:

Hello,

In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:

$n_{SF_6}=10.0gSF_6*\frac{1molSF_6}{146.06 gSF_6}=0.0685mol$

Next, for the given energy, we compute the total heat that is liberated:

$Q=-1220.47\frac{kJ}{mol}*0.0685 mol\\\\Q=-76.7kJ$

Finally, we conclude such symbol has sense since negative heat is related with liberated heat.

Best regards.

8 0

3 years ago

Consider the molecules: scl2, f2, cs2, cf4, and brcl. select the choice where the bonds of the molecules are arranged in order o

elena-s [515]

The polarity of a bond depends on the difference in electronegativities of the atoms involved in bonding. Greater this difference, greater is the polarity

1) In SCl2 we have the S-Cl bond

Electronegativity values: S = 2.5, Cl = 3

Difference = 3.0 -2.5 = 0.5

2) In F2 we have the F-F bond

Electronegativity value: F = 4.0

Difference = 4.0-4.0 = 0

3) In CS2 we have the C-S bond

Electronegativity value: C = 2.5, S = 2.5

Difference = 2.5 -2.5 = 0

4) In CF4 we have the C-F bond

Electronegativity values: C = 2.5, F = 4.0

Difference = 4.0 -2.5 = 1.5

5) In BrCl we have the Br-Cl bond

Electronegativity values: Br = 2.8, Cl = 3

Difference = 3.0 -2.8 = 0.2

Therefore, the increasing order of polarity is

F2 = CS2 < BrCl < SCl2 < CF4

5 0

3 years ago

Rank the following elements in order of decreasing atomic radius (1 is biggest, 5 is smallest)

AlexFokin [52]

There are various kind of elements that are present in periodic table. Some elements are harmful, some are radioactive, some are noble gases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

<h3>What is periodic table?</h3>

Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.

Along the period, the size of elements decreases. Down the group the size of elements increases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

Therefore, atomic radius in decreasing order is Bi>Sb>As>N>O.

Learn more about periodic table, here:

brainly.com/question/11155928

#SPJ1

3 0

1 year ago

Can someone please list one of the groups of representative elements in 1A-7A

Vadim26 [7]

Group 1A(1), the alkali metals, includes lithium, sodium, and potassium. Group 7A(17) the halogens, includes chlorine, bromine, and iodine. hope this helps:)

7 0

2 years ago

Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the

expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

Volume of $Ca(OH)_2$ = 14.5 mL

Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

So, in 14.5mL. the mass of $Ca(OH)_2$ present will be = $\frac{0.185}{100}\times 14.5=0.0268g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = $2.50\times 10^{-3}$

Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

5 0

3 years ago

Which one would it be 1,2,3,4