Chemistry, 28.12.2019 02:31, raman2003

Consider the following standard heats of formation:
p₄o₁₀(s) = -3110 kj/mol
h₂o(l) = -286 kj/mol
h₃po₄(s) = -1279 kj/mol
calculate the change in enthalpy for the following process:
$p_4o_{10}(s) + 6h_2o(l) \rightarrow 4h_3po_4(s)$

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answered: nate1712

The Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = -290\,kJ$

Explanation:

Given- Standard Heat of Formation:

$\Delta H_{f}^{\circ } [P_{4}O_{10}(s)]$ = -3110 kJ/mol,

$\Delta H_{f}^{\circ } [H_{2}O(l)]$ = -286 kJ/mol,

$\Delta H_{f}^{\circ } [H_{3}PO_{4}(s)]$ = -1279 kJ/mol

Given chemical reaction: P₄O₁₀(s) + 6H₂O → 4H₃PO₄

The standard enthalpy of reaction: $\Delta H_{r}^{\circ }$ = ?

To calculate the Standard enthalpy of reaction ( $\Delta H_{r}^{\circ }$ ), we use the equation:

$\Delta H_{r}^{\circ } = \sum \nu .\Delta H_{f}^{\circ }(products)-\sum \nu .\Delta H_{f}^{\circ }(reactants)$

$\Delta H_{r}^{\circ } = [4 \times \Delta H_{f}^{\circ } [H_{3}PO_{4}(s)]] - [1 \times \Delta H_{f}^{\circ } [P_{4}O_{10}(s)] + 6 \times \Delta H_{f}^{\circ } [H_{2}O(l)]]$

$\Rightarrow \Delta H_{r}^{\circ } = [4 \times (-1279\, kJ/mol)] - [1 \times (-3110\, kJ/mol) + 6 \times (-286\, kJ/mol)]$

$\Rightarrow \Delta H_{r}^{\circ } = [-5116\, kJ] - [-3110\, kJ -1716\, kJ]$

$\Rightarrow \Delta H_{r}^{\circ } = [-5116\, kJ] - [-4826\, kJ] = -290\,kJ$

Therefore, the Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = -290\,kJ$

answered: Guest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

answered: Guest

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o