How much carbon dioxide (CO2). Is in our atmosphere?

Chemistry

1 answer:

Brut [27]3 years ago

6 0

: Atmospheric Carbon Dioxide. The global average atmospheric carbon dioxide in 2017 was 405.0 parts per million (ppm for short), with a range of uncertainty of plus or minus 0.1 ppm. Carbon dioxide levels today are higher than at any point in at least the past 800,000 years.

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Scientists are studying cell-to-cell communication in a human spleen.

Nady [450]

Answer:

To predict what triggers cell-to-cell communication

Explanation:

Methodologies used for studying cell-to-cell communication includes

1) The reductionist method in which cellular response are measured based on testing specific cellular signals by the introduction of different chemicals, so as to find the source that triggers the response

2) By the application of known chemical reagents enable or block specific areas of an organism's receptors and with the aid of an electrode measure the cellular response based on generated currents of ions or by collecting and analyzing fluid samples from the affected area.

7 0

3 years ago

Set up a Hess’s law cycle, and use the following information to calculate ΔH∘f for aqueous nitric acid, HNO3(aq). You will need

Assoli18 [71]

Answer:

$\Delta H_{f}^{\circ } (HNO_{3}, aq)$ = (-207.39) kJ/mol

Explanation:

Given:

$\Delta H_{f}^{\circ } (NH_{3}, g)= -46.1 kJ/mol\\ \Delta H_{f}^{\circ } (H_{2}O, l)= -285.8 kJ/mol\\ \Delta H_{f}^{\circ } (HNO_{3}, aq)= ? kJ/mol$

3NO₂(g)+H₂O(l)→ 2HNO₃(aq)+NO(g); ΔH° = -137.3 kJ ....equation 1

2NO(g)+O₂(g)→ 2NO₂(g) ; ΔH° = -116.2 kJ ....equation 2

4NH₃(g)+5O₂(g)→ 4NO(g)+6H₂O(l) ;ΔH° = -1165.2 kJ ....equation 3

Multiplying equation 1 with (2/3), we get

2NO₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) + (2/3) NO(g) ....equation 4

Adding equation 2 and 4, we get

2NO₂(g) + 2/3 H₂O(l) + 2NO(g) + O₂(g) → 4/3 HNO₃(aq) + (2/3) NO(g) + 2NO₂(g)

⇒ (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) ....equation 5

Multiplying equation 3 with (1/3), we get

(4/3) NH₃(g) + (5/3) O₂(g) → (4/3) NO(g)+ 2 H₂O(l) ....equation 6

Now adding equation 5 and 6, we get

(4/3) NH₃(g) + (5/3) O₂(g) + (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → (4/3) NO(g)+ 2 H₂O(l) + 4/3 HNO₃(aq)

⇒ (4/3) NH₃(g) + (8/3) O₂(g) → (4/3) HNO₃(aq) + (4/3) H₂O(l) ....equation 7

Now multiplying equation 7 with (3/4), we get

NH₃(g) + 2 O₂(g) → HNO₃(aq) + H₂O(l) ...equation 8

Therefore, by Hess's law the standard enthalpy of formation is:

ΔH° = (3/4) [ (-137.3 kJ) × (2/3) + (-116.2 kJ) + (-1165.2 kJ) × (1/3)]

ΔH° = (3/4) [ - 91.53 - 116.2 - 388.4]

ΔH° = (3/4) [-596.13] = -447.09 kJ

Since the change in enthalpy of a reaction:

ΔH° = $[\Delta H_{f}^{\circ } (HNO_{3}, aq) + \Delta H_{f}^{\circ } (H_{2}O, l)] - [\Delta H_{f}^{\circ } (NH_{3}, g) + 2 \Delta H_{f}^{\circ } (O_{2}, g)]$