Answer:
Threatened Species: A threatened species is a species at risk but not yet endangered. California sea otters were classififed as a threatened species. Laws were passed to protect the otters and now they have increased their population size.
Invasive Species: One of the main causes of extinction and endangered species is the introduction of an exotic species. New exotic species are called invasive species. Invasive species can disrupt food chains, carry disease, prey on native species directly, and out-compete native species for limited resources, like food.
Extinction: If a population decreases too much in numbers, they disappear. Extinct species mean that the species has died out and no individuals left. An example of extinction: New Zealand was once home to a bird called the Giant Moa. Humans settled as their population increased the Moa population decreased. The species is now extinct.
Explanation:
Answer:
BaSO₄
Explanation:
Let's assume we have 100 g of the compound. If that's the case we would have:
Now we <u>convert the masses of each element into moles</u>, using their <em>respective molar masses</em>:
- 58.8 g Ba ÷ 137.327 g/mol = 0.428 mol Ba
- 13.74 g S ÷ 32 g/mol = 0.429 mol S
- 27.43 g O ÷ 16 g/mol = 1.71 mol O
We <u>divide those moles by the lowest number among them</u>:
We can express those results as Ba₁S₁O₄, meaning the empirical formula is thus BaSO₄.
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
