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2 years ago

Over-the-counter hydrogen peroxide solutions are 3% (w/v). What is this concentration in moles per liter?

1 answer:

5 0

1 ml of water weighs 1 gram so 1 liter (1000 ml) weighs 1000 grams. A 3% solution (3% = 0.03) of hydrogen peroxide (w/v) would contain 1000 grams x 0.03 or 30 grams. The chemical formula of hydrogen peroxide is H2O2 and a mole weighs 34.0147 grams/mole. So 30 grams of H2O2 divided by 34.0147 grams/mole equals 0.88 moles of H2O2. The concentration of a 3% (w/v) hydrogen peroxide solution therefore contains 30 grams of H202 (or 0.88 moles of H202) per in a liter of water (or 1000 grams H20) would thus be 0.88 moles H2O2 per liter (0.88 moles H2O2/l) .

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g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole

Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

3 0

2 years ago

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0

2 years ago

A salt solution will conduct electricity but if sugar solution will not explain why

Rainbow [258]

Salt solution such as sodium chloride (NaCl) conducts an electric current because it has ions in it that have the freedom to move about in solution. ... On the other hand, sugar solution does not conduct an electric current because sugar (C12H22O11) dissolves in water to produce sugar molecules

6 0

2 years ago

Having studied the data for patterns and trends, you should now be able to draw

avanturin [10]

Answer:

The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock.

Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.

The time elapsed will be log (L) atoms X half-life.

Explanation:

3 0

2 years ago

A glucose solution has a density of 1.02 g/ml. what is its specific gravity

Makovka662 [10]

Answer is: specific gravity of glucose is 1,02.
d(glucose) = 1,02 g/ml.
d(water) = 1,00 g/ml.
Specific gravity of glucose = density of glucose ÷ density of water.
Specific gravity of glucose = 1,02 g/ml ÷ 1,00 g/ml.
Specific gravity of glucose = 1,02.
Specific gravity is the ratio of the density of a substance (in this case glucose) to the density of a reference substance (water).

4 0

2 years ago

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