Answer:
55.0 g
Explanation:
Step 1: Write the balanced equation for the production of oxygen
2 CO₂(g)⇒ 2 CO(g) + O₂(g)
Step 2: Calculate the mass of oxygen produced over a 2 hour period
The MOXIE produces 10.0 g of oxygen per hour.
2 h × 10.0 g/1 h = 20.0 g
Step 3: Calculate the moles corresponding to 20.0 g of O₂
The molar mass of O₂ is 32.00 g/mol.
20.0 g × 1 mol/32.00 g = 0.625 mol
Step 4: Calculate the number of moles of CO₂ needed to produce 0.625 moles of O₂
The molar ratio of CO₂ to O₂ is 2:1. The moles of CO₂ needed are 2/1 × 0.625 mol = 1.25 mol
Step 5: Calculate the mass corresponding to 1.25 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
1.25 mol × 44.01 g/mol = 55.0 g
The criteria for designation as a biodiversity hotspot include species richness, endemism, and risk of extinction.
<h3>What is a biodiversity hotspot?</h3>
A biodiversity hotspot is a region having high levels of biodiversity in its communities and ecosystems.
A biodiversity hotspot must have a high number of endemic vascular plants (at least 1500 species).
In conclusion, the criteria for designation as a biodiversity hotspot include species richness, endemism, and risk of extinction.
Learn more about biodiversity hotspots here:
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"It is important to balance chemical equations because there must be an equal number of atoms on both sides of the equation to follow the Law of the Conservation of Mass"
Answer:In alpha decay, shown in Fig. 3-3, the nucleus emits a 4He nucleus, an alpha particle. Alpha decay occurs most often in massive nuclei that have too large a proton to neutron ratio. An alpha particle, with its two protons and two neutrons, is a very stable configuration of particles. Alpha radiation reduces the ratio of protons to neutrons in the parent nucleus, bringing it to a more stable configuration. Many nuclei more massive than lead decay by this method.
Consider the example of 210Po decaying by the emission of an alpha particle. The reaction can be written 210Po Æ 206Pb + 4He. This polonium nucleus has 84 protons and 126 neutrons. The ratio of protons to neutrons is Z/N = 84/126, or 0.667. A 206Pb nucleus has 82 protons and 124 neutrons, which gives a ratio of 82/124, or 0.661. This small change in the Z/N ratio is enough to put the nucleus into a more stable state, and as shown in Fig. 3-4, brings the "daughter" nucleus (decay product) into the region of stable nuclei in the Chart of the Nuclides.
In alpha decay, the atomic number changes, so the original (or parent) atoms and the decay-product (or daughter) atoms are different elements and therefore have different chemical properties.
Upper end of the Chart of the Nuclides
In the alpha decay of a nucleus, the change in binding energy appears as the kinetic energy of the alpha particle and the daughter nucleus. Because this energy must be shared between these two particles, and because the alpha particle and daughter nucleus must have equal and opposite momenta, the emitted alpha particle and recoiling nucleus will each have a well-defined energy after the decay. Because of its smaller mass, most of the kinetic energy goes to the alpha particle.