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Digiron [165]
3 years ago
12

Identify the combustion reaction. identify the combustion reaction. c(s) + o2(g) → co2(g) 2 h2(g) + o2(g) → 2 h2o(g) c3h8(g) + 5

o2(g) → 3 co2(g) + 4 h2o(g) 2 c3h7oh(l) + 9 o2(g) → 6 co2(g) + 8 h2o(g) all of the reactions are examples of combustion reactions.
Chemistry
1 answer:
tester [92]3 years ago
7 0
Combustion reaction occurs when organic compound reacts with oxygen to form CO₂, H₂O and energy
C(s) + O₂(g) → CO₂(g) is Synthesis reaction not combustion
2 H₂(g) + O₂(g) → 2 H₂O(g) also Synthesis reaction and not combustion
C₃H₈(g) + 5 O₂ → 3 CO₂(g) + 4 H₂O(g) is considered as Combustion reaction
2 C₃H₇OH(l) + 9 O₂(g) → 6 CO₂(g) + 8 H₂O(g) Combustion reaction
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Identify the acid and conjugate base pair in the following equation:
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7 0
3 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0
3 years ago
4. Why can't the subscripts be changed in a chemical equation in chemistry
neonofarm [45]
If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is
4 0
3 years ago
How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102
Ede4ka [16]

Answer: 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles

1 mole of hydrogen (H_2) = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms

17.5 mole of hydrogen (H_2) = \frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25} atoms

There are 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0
3 years ago
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