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miss Akunina [59]
3 years ago
10

Balance the equation αC6H14(g)+βO2(g)→γCO2(g)+δH2O(g)

Chemistry
2 answers:
fenix001 [56]3 years ago
7 0
Answer: The balanced reaction is:
<span>2 C6H14(g)+ 19 O2(g) → 12 <span>CO2</span>(g)+ 14 H2O(g)

Note: While balancing the chemical reaction, care must be taken that total number atoms (of each type) on both reactant and product side must be same. In present case, there are 12 'C' atoms, 28 'H' atoms and 38 'O' atoms on both reactant and product side. Hence, the reaction is balanced. </span>
Leona [35]3 years ago
3 0
2 \ C_6H_{14}_{(g)} + 19 \ O_2_{(g)} \rightarrow 12 \ CO_2_{(g)} + 14 \ H_2O_{(g)}
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A 20.0 mL solution of NaOH is neutralized with 24.1 mL of 0.200 M HBr. What is the concentration of the original NaOH solution
Alinara [238K]

Answer:

0.241 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HBr + NaOH —> NaBr + H₂O

From the balanced equation above,

The mole ratio of acid, HBr (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 1

Finally, we shall determine the concentration of the NaOH solution. This can be obtained as follow:

Volume of base, NaOH (V₆) = 20 mL

Volume of acid, HBr (Vₐ) = 24.1 mL

Concentration of acid, HBr (Cₐ) = 0.2 M

Concentration of base, NaOH (C₆) =?

CₐVₐ / C₆V₆ = nₐ/n₆

0.2 × 24.1 / C₆ × 20 = 1/1

4.82 / C₆ × 20 = 1

Cross multiply

C₆ × 20 = 4.82

Divide both side by 20

C₆ = 4.82 / 20

C₆ = 0.241 M

Therefore, the concentration of the NaOH solution is 0.241 M

8 0
3 years ago
How many water molecules self ionize in one liter of water
Drupady [299]

Answer:

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Explanation:

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10. When a 13.6 g sample of a compound containing only magnesium and oxygen is decomposed, 5.4 g of oxygen is obtained. What is
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3 years ago
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The normal boiling point of iodomethane, CH3I, is 42.43 8C, and its vapor pressure at 0.00 8C is 140. Torr. Calculate (a) the st
tigry1 [53]

Answer:

a=28600J; b=90.6 J/K; c=402 torr

Explanation:

(a) considering the data given

 Vapour pressure P1 =0  at Temperature T1 = 42.43˚C,

Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)

Using the Clausius-Clapeyron Equation

ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

In 760/140 = ΔH/8.314 J/mol/K  × (1/315.58K -- 1/273.15K)

ΔH vap= +28.6 kJ/mol or 28600J

(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.

Since ΔG at boiling point is zero,

ΔS =(ΔH°vap/Τb)

 ΔS = 28600 J/315.58 K

= 90.6 J/K

(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

ln P298 K/1 atm =  28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)

P298 K = 0.529 atm

                = 402 torr

8 0
3 years ago
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