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swat32
3 years ago
9

The mass of the erlenmeyer flask alone is 88.000 grams. calculate the mass of the zinc and hcl after the addition of 25 grams of

zinc to an open flask containing 15 ml hcl.
Chemistry
1 answer:
arsen [322]3 years ago
7 0
If concentration of HCl is 1 mol/dm³ :
m(<span>erlenmeyer flask) = 88,00 g.
m(Zn) = 25,0 g.
V(HCl) = 15 ml = 15 cm</span>³ = 0,015 dm³.
Chemical reaction: Zn + 2HCl → ZnCl₂ + H₂.
n(HCl) = c(HCl) · V(HCl).
n(HCl) = 1 mol/dm³ · 0,015 dm³ = 0,015 dm³.
n(Zn) = 25 g ÷ 65,4 g/mol = 0,38 mol.
n(H₂) = 0,015 mol ÷ 2 = 0,0075 mol.
m(H₂) = 0,0075 mol · 2g/mol =  0,015 g.

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4 0
3 years ago
The mass of Carbon is 12 g/mol and the mass of oxygen is 16 g/mol. What is the approximate percent composition by mass of CO2 (f
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the last one

27% by 73%

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12/44*100 for C

and

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5 0
3 years ago
1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
nikklg [1K]

Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

Best regards.

4 0
3 years ago
A 25.00 mL sample of the ammonia solution
musickatia [10]

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

3 0
3 years ago
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