Did you mean moles? Instead of miles
If so it’s 0. 330568289543697
Answer:
ΔH°(f) = -110.5 Kj/mole (exothermic)
Explanation:
C + 1/2O₂ => CO
This is asking for the 'Standard Heat of Formation (ΔH°(f)* for carbon monoxide (CO). Values for many compounds can be found in the appendix of most college general chemistry text books. From Ebbing & Gammon, 11th edition, General Chemistry, Appendix C, page 8A.
*Standard Heat of Formation by definition is the heat gained or lost on formation of a substance (compound) from its basic elements in standard state.
The ΔH°(f) values as indicated are found in the appendix of most college chemistry texts. By choosing any compound, one can determine the standard heat of formation equation for the substance of interest. For example, consider Magnesium Carbonate; MgCO₃(s).The basic standard states of each element is found in the Appendix on Thermodynamic Properties for Substances at 25°C & 1 atm. having ΔH°(f) values = 0.00 Kj/mole. All elements in standard state have a 0 Kj/mol. See appendix and note that under the ΔH°(f) symbol some substances have 0.00 Kj/mol values. The associated element will be in basic standard state,
Standard Heat of Formation Equation for formation of Magnesium Carbonate;
Mg°(s) + C°(gpt)* + 3/2O₂(g) => MgCO₃(s) ; ΔH°(f) = -1111.7 Kj/mole
* gpt => graphite
Answer:
0.46M NaS₂O₃ (Assuming KIO₃ solution with a concentration of 1.0M)
Explanation:
Based on the reaction:
6 Na₂S₂O₃ + KIO₃ + 5 KI + 3 H₂SO₄ → 3 Na₂S₄O₆ + 3 H₂O + 3 K₂SO₄ + 6 NaI
<em>6 moles of Na₂S₂O₃ react per mole of KIO₃</em>
Assuming the molarity of the KIO₃ solution is 0,1M:
Moles of KIO₃: = 5.0x10⁻³L ₓ (0.1 mol / L) = <em>5.0x10⁻⁴ moles</em>
As 6 moles of thiosulfate reacted per mole of iodate:
5.0x10⁻⁴ moles KIO₃ ₓ (6 moles Na₂S₂O₃ / 1 mole KIO₃) =
<em>3.0x10⁻³ moles of Na₂S₂O₃. </em>In 6.5mL (6.5x10⁻³L):
3.0x10⁻³moles Na₂S₂O₃ / 6.5x10⁻³ L = 0.46M NaS₂O₃
The molarity of the hydrochloric acid, HCl solution given the data is 0.079 M
<h3>Balanced equation </h3>
HCl + NaOH —> NaCl + H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the molarity of HCl </h3>
- Molarity of base, NaOH (Mb) = 0.137 M
- Volume of base, NaOH (Vb) = 16.1 mL
- Volume of acid, HCl (Va) = 27.9 mL
- Molarity of acid, HCl (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 27.9) / (0.137 × 16.1) = 1
(Ma × 27.9) / 2.2057 = 1
Cross multiply
Ma × 27.9 = 2.2057
Divide both side by 27.9
Ma = 2.2057 / 27.9
Ma = 0.079 M
Thus, the molarity of the HCl solution is 0.079 M
Learn more about titration:
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Because it contain very high carbon contend as compare to paper
