How do I write the following elements:

Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc

Ber [7]

Answer:

The balanced equation for this reaction will be

$CH4 + 4F2$ → $CF4 + 4HF$

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

1 mole of CH4 → 4 mole of 4 mole of fluorine

0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

4 moles of fluorine → 1 mole of CF4

0.56 mole → $\frac{1}{4} * 0.56$ = 0.14mole of CF4

4 moles of fluorine → 4 moles of HF

0.56 mole of fluorine → 0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0

3 years ago

Why the partial charge on H in HF is more positive than the partial charge on H in HBr

erik [133]

Because fluorine has a higher electronegativity

8 0

3 years ago

Calculate the molarity of sodium ion in a solution made

Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Sodium chloride will furnish Sodium ions as:

$NaCl\rightarrow Na^{+}+Cl^-$

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

$Moles =0.288 \times {3.58\times 10^{-3}}\ moles$

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

$Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}$

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

$Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles$

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}$

<u> The final concentration of sodium anion = 0.1035 M</u>

4 0

3 years ago

Choose all the answers that apply. Sound _____. travels in transverse waves requires a medium to travel through travels most slo

kari74 [83]

Answer: Option (a) is the correct option.

Explanation:

In order to travel, sound does need a medium. The more closer the particles of a medium, the better sound can travel through it. Therefore, in solids sound travels very quickly because in solids particles are closer to each other. As a result, they can easily transmit the energy from one particle to another.

Thus, it becomes easier for sound to travel through it. On the other hand, in solids sound can travel in both longitudinal and transverse waves.

Hence, out of the given options, sound travels in transverse waves requires a medium to travel through is the correct option.

3 0

3 years ago

Consider conjugation. A long molecule contains a conjugated network of double bonds. Two of these double bonds begin at carbon 2

kaheart [24]

Answer: 4 and 6

Explanation: the double bonds are alternating in the chain

C—C=C—C=C—C=C—C=

1 2 3 4 5 6 7 8

3 0

2 years ago