Answer:
The answer to your question is 432 g of CO₂
Explanation:
Data
CaCO₃ = 983 g
CaO = 551 g
CO₂ = ?
Balanced reaction
CaCO₃ (s) ⇒ CaO (s) + CO₂ (g)
This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.
Mass of reactants = Mass of products
Mass of CaCO₃ = Mass of CaO + Mass of CO₂
Solve for CO₂
Mass of CO₂ = Mass of CaCO₃ - Mass of CaO
Mass of CO₂ = 983 g - 551 g
Simplification
Mass of CO₂ = 432 g
The ecosystem services provided by the each of the given examples are as follows:
- A cornfield in Kansas provides provisioning services.
- Bacteria that decompose waste along the Gulf Coast provides regulating services.
- Ocean currents that keep Pacific Northwest air cool and moist provides regulating services.
- Flower garden at a national landmark provides cultural services.
- Lumber from an oak tree provides provisioning services.
- Animals that eat seeds and then spread the seeds through their waste regulating services.
<h3>Ecosystem services</h3>
Ecosystem services are defined as the benefits derived by man from the surroundings ecosystems.
<h3>Categories of ecosystem services</h3>
The four categories of ecosystem services are:
- regulating services,
- provisioning services,
- cultural services, and
- supporting services
A cornfield in Kansas provides provisioning services.
Bacteria that decompose waste along the Gulf Coast provides regulating services.
Ocean currents that keep Pacific Northwest air cool and moist provides regulating services.
Flower garden at a national landmark provides cultural services.
Lumber from an oak tree provides provisioning services.
Animals that eat seeds and then spread the seeds through their waste regulating services.
Learn more about ecosystem services at: brainly.com/question/2191258
Answer:
b) Delta S < 0
Explanation:
The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).
- If Δn(g) > 0, the entropy increases (ΔS > 0).
- If Δn(g) < 0, the entropy decreases (ΔS < 0).
- If Δn(g) = 0, there is little or no change in the entropy
Let's consider the following equation.
2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)
Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.
Answer:
Vertically Shrunk by a factor of 1/6
Explanation:
Parent Formula: f(x) = a(bx - c) + d
<em>a</em> - vertical shrink/stretch and x-reflections
<em>b</em> - horizontal shrink/stretch and y-reflections
<em>c</em> - horizontal movement left/right
<em>d</em> - vertical movement up/down
Since we are only modifying <em>a</em>, we are dealing with vertical shrink/stretch:
Since a < 1 (1/6 < 1), we are dealing with a vertical shrink of 1/6.
Since a > 0 (1/6 > 0), we do not have a reflection over the x-axis.
Use a periodic table, It should help
