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stellarik [79]
3 years ago
9

Use enthalpies of formation given in appendix c to calculate δh for the reaction br2(g)→2br(g), and use this value to estimate t

he bond enthalpy d(br−br).
Chemistry
1 answer:
Contact [7]3 years ago
3 0

Given reaction represents dissociation of bromine gas to form bromine atoms

Br2(g) ↔ 2Br(g)

The enthalpy of the above reaction is given as:

ΔH = ∑n(products)ΔH^{0}f(products) - ∑n(reactants)ΔH^{0}f(reactants)

where n = number of moles

ΔH^{0}f= enthalpy of formation

ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol

Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol

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A waste collection tank can hold 18,754 kg of methanol, which has a density of 0.788 g/cm3. What is the volume of the waste coll
Ivahew [28]
A is correct hope this helps
4 0
3 years ago
Read 2 more answers
The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.
Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0
3 years ago
To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w
Elis [28]

<u>Answer:</u> The number of moles of weak acid is 4.24\times 10^{-3} moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}

We are given:

Volume of solution = 43.81 mL = 0.04381 L      (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol

The chemical reaction of weak monoprotic acid and KOH follows the equation:

HA+KOH\rightarrow KA+H_2O

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, 4.24\times 10^{-3}mol of KOH will react with = \frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol of weak monoprotic acid.

Hence, the number of moles of weak acid is 4.24\times 10^{-3} moles.

6 0
3 years ago
How many liters of 1.75 M solution could be made using 35 grams of NaCl?
dolphi86 [110]
Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
1.75 =  \frac{35}{58.44*V}
1.75*58.44V = 35
102.27V = 35
V =  \frac{35}{102.27}
\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
Hydrofluoric acid, hf(aq), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the
Trava [24]
1) Chemical equation

Na2 SiO3 (s) + 8 HF (aq) ---> H2 Si F6 (aq) + 2 Na F (aq) + 3H2O (l)

It is balanced

2) Molar ratios

1 mol Na2 SiO3 : 8 mol HF.

3) Proportion

0.340 mol Na2 SiO3 * 8 mol HF / 1mol Na2SiO3 = 2.72 mol HF.

Answer: 2.72 mol HF

5 0
4 years ago
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