Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, since the potassium bromide and silver nitrate are in a 1:1 mole ratio, the first step is to identify the limiting reactant, by considering the reacting volumes of reactants in order to compute the available moles of potassium bromide and the moles of potassium bromide consumed by the 15.4 mL of 0.512-M solution of silver nitrate:
In such a way, since less moles are consumed than available, we infer that silver nitrate is the limiting reactant, for which the resulting grams of silver bromide (molar mass 187.8 g/mol) result:
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Answer:
There are three amino acids that have basic side chains at neutral pH. These are arginine (Arg), lysine (Lys), and histidine (His). Their side chains contain nitrogen and resemble ammonia, which is a base. Their pKa's are high enough that they tend to bind protons, gaining a positive charge in the process.
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The answer you're looking for is photosynthesis
Answer:
The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.
or the
FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.
Element Atomic number Electron configuration
sulfur 16 1s22s22p63s23p4
chlorine 17 1s22s22p63s23p5
argon 18 1s22s22p63s23p6
or the
Two electrons
Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.
Explanation:
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Answer:
Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:
So we solve for C2:
2. Now, since 111 mL of water is added, we compute the final volume, V3:
So, the final concentration of the 139 mL portion is:
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