Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
To solve for the number of moles, we simply have to use the Avogadros number which states that there are 6.022 x 10^23 molecules per mole. Therefore:
number of moles = 6.67 X 10^40 chlorine molecules / (6.022 x 10^23 molecules / mole)
number of moles = 1.108 x 10^17 moles
Answer:
B protons determine indentity and valence electrons determine chemical properties.
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Explanation:
1A: The legs can be a adjusted, as well as the sand can be swapped out. It’s a very good design for running multiple tests.
1B: He could add books or something under the front or back legs in order to increase/decrease the incline, therefore imitating the hypothesis.
1C: He can change out the sand grains to finer ones, or coarser ones, and record his results of each test.
2: If he sets the model at a steep incline and tests it with coarse sand and fine sand, seeing which one makes a narrower, deeper hole.
Answer:
B. The collisions release heat, which results in the heating and subsequent melting, sinking, and rising of materials.
Explanation:
Got it correct on edge