I’m going to say the answer is most likely X because memetic energy is energy being used at that moment. Because the coaster is going fastest at X we can assume that the answer is X
Answer:
Mass of sea food = 30.98 Kg
Mass of sea food in pound = 68.31 lbs
Explanation:
Salmon, crab and oysters all are sea food.
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
Mass of salmon = 22 kg
Mass of crab = 5.5 kg
Mass of oysters = 3.48 kg
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
= 22 + 5.5 + 3.48
= 30.98 Kg
1 Kg = 2.205 lbs
Therefore, 30.98 kg = 30.98 × 2.205
= 68.31 lbs
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
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Answer:
Samira's model correctly demonstrates how the properties changed with the rearrangement of the atoms. However not all atoms are accounted for. There is a missing reactant.
Explanation:
Samira's model correctly demonstrated how the atoms in two compounds reacted to form two new products. However, the elements present in the reactants side should be the elements that make up the new products in the product side. But as the diagram shows, Sameera has mistakenly added a new element to one of her products which will be wrong.
