Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
DO you have any clues that may help to solve this?
Answer:
Rate = 43 M⁻¹s⁻¹[NO₂][O₃]
Explanation:
We need to find the reaction order in
rate = (NO₂ )ᵃ (O₃ )ᵇ
given:
( NO₂ ) M ( O₃ ) M Rate M/s
0.10 0.33 1.420 (1)
0.10 0.66 2.840 (2)
0.25 0.66 7.10 (3)
When keeping the NO₂ concentration constant in the first two while doubling the concentration of O₃ , the rate doubles. Therefore it is first order with respect to O₃
Comparing (2) and (3) increasing the concentration of NO₂ by a factor of 2.5 and keeping O₃ constant , increased the rate by a factor of 2.5. Therefore the rate is first order with respect to NO₂
Then rate law is
= k (NO₂) (O₃ )
To find k take any of the three and substitute the values to find k:
1.420 M/s = k (0.10)M x (0.33)M ⇒ k = 43 /Ms
Then the answer is Rate = 43 M⁻¹s⁻¹[NO₂][O₃]
Answer:
The word gentle traces back to the Latin word gentlis, meaning “of the same clan,” and at first the world was used to describe people belonging to distinguished families, who were seen as courteous and noble. Nowadays you’re more likely to hear the word used to refer to things that are calm, moderate, and without harshness. The word is often applied to people, but it can be used more broadly to describe anything that is mild, such as "a gentle detergent" or "a gentle reminder."
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
