When aqueous ammonia is added to the half-cell containing 0.001 M CuSO4, the cell potential is likely to change. The reason for this is that ammonia can form coordination complexes with copper ions, which can affect the concentration of copper ions in the solution, and hence the concentration gradient that drives the redox reaction in the cell.
Ammonia can react with copper ions in aqueous solution to form a series of coordination complexes. The most common complex is Cu(NH3)42+, which is a tetraamminecopper(II) complex. The formation of this complex reduces the concentration of free Cu2+ ions in solution, which can shift the equilibrium of the redox reaction in the cell.
If the reduction half-reaction is Cu2+ + 2e- → Cu, the addition of ammonia can reduce the concentration of Cu2+ ions in the solution and shift the equilibrium to the left, decreasing the cell potential. On the other hand, if the oxidation half-reaction is Cu → Cu2+ + 2e-, the addition of ammonia can increase the concentration of Cu2+ ions and shift the equilibrium to the right, increasing the cell potential.
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The cloud droplets in a cloud are formed by water vapor molecules and: A) protons. B) ions. C) molecules of air. D) condensation nuclei.
Answer:
condensation nuclei
Explanation:
consider the reaction performed in the sn1 lab. what would be the effect on the rate of the reaction if 2-propanol (isopropanol) was used instead of 2-methyl-2-propanol (t-butanol) assuming only an sn1 reaction occurs? group of answer choices the rate of the reaction would decrease, because the secondary carbocation is more difficult to form. the rate of the reaction would increase, because the secondary carbocation is easier to form. there would be no difference in reaction rate. the reaction would not proceed at all.
The rate of the reaction is directly proportional to the stability of the carbocation intermediate, and any changes in the solvent will affect the rate of the reaction.
In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate. The stability of the carbocation intermediate affects the rate of the reaction.
In this case, if 2-propanol (isopropanol) was used instead of 2-methyl-2-propanol (t-butanol), the rate of the reaction would decrease. This is because the carbocation intermediate formed in 2-propanol is less stable compared to the one formed in t-butanol.
The carbocation intermediate formed in t-butanol is tertiary, which is more stable than the one formed in isopropanol, which is secondary. This means that the reaction will be slower in isopropanol due to the less stable carbocation intermediate.
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What is the concentration (in molality) of an aqueous solution of NaCl made by adding
4.56 g of NaCl to enough water to give 20.0 mL of solution. Assume the density of the
solution is 1.03 g/mL
Answer:
data given
mass of NaCl 4.56
dissolved volume 20ml(0.02l)
density of solution 1.03g/ml
Required molality
Explanation:
molarity=m/mr×v
where
m is mass
mr molar mass
v is volume
now,
molarity=4.56/58.5×0.02
molarity =3.9
: .molarity is 3.9mol/dm^3
According to molal concentration, the concentration (in molality) of an aqueous solution of NaCl is 0.0047 mole/kg.
What is molal concentration?Molal concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molal concentration is moles/kg.
The molal concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molal concentration is calculated by the formula, molal concentration=mass/ molar mass ×1/mass of solvent in kg.
In terms of moles, it's formula is given as molal concentration= number of moles /mass of solvent in kg.
Substitution in formula gives the answer but first mass of solution is determined which is density×volume= 1.03×20=20.6 g , mass of solvent= 20.6-4.56=16.05, thus molal concentration=4.56/58.5×1/16.05=0.0047 moles/kg.
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PLEASE ANSWER ASAP
1. How many atoms are present in 8.500 mole of chlorine atoms?
2. Determine the mass (g) of 15.50 mole of oxygen.
3. Determine the number of moles of helium in 1.953 x 108 g of helium.
4. Calculate the number of atoms in 147.82 g of sulfur.
5. Determine the molar mass of Co.
6. Determine the formula mass of Ca3(PO4)2.
IT WOULD BE HELPFUL
1) 5.1167 x 10²⁴atoms of chlorine. 2) 248.00 g. 3) 4.8825 x 10⁷ moles of helium. 4) 2.7757 x 10²⁴ atoms of sulfur. 5) Molar mass of Co (cobalt) is 58.93 g/mol. 6) Formula mass = 310.18 g/mol.
What is meant by formula mass?Sum of the atomic masses of all the atoms in chemical formula is called formula mass
1.) Number of atoms = 8.500 moles x 6.022 x 10²³ atoms/mole = 5.1167 x 10²⁴ atoms of chlorine.
2.) Molar mass of oxygen is 16.00 g/mol. Therefore:
Mass of 15.50 moles of oxygen = 15.50 moles x 16.00 g/mol = 248.00 g.
3.) Molar mass of helium is 4.00 g/mol. Therefore, the number of moles of helium in 1.953 x 10⁸ g is:
Number of moles = 1.953 x 10⁸ g / 4.00 g/mol = 4.8825 x 10⁷ moles of helium.
4.) Molar mass of sulfur is 32.06 g/mol. Therefore, the number of moles of sulfur in 147.82 g is:
Number of moles = 147.82 g / 32.06 g/mol = 4.6084 moles of sulfur.
To find the number of atoms, we can use Avogadro's number again:
Number of atoms = 4.6084 moles x 6.022 x 10²³ atoms/mole = 2.7757 x 10²⁴ atoms of sulfur.
5.) Molar mass of Co (cobalt) is 58.93 g/mol.
6.) Ca₃(PO₄)₂ contains 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms.
Atomic masses of these elements are:
Calcium (Ca) = 40.08 g/mol
Phosphorus (P) = 30.97 g/mol
Oxygen (O) = 16.00 g/mol
Therefore, formula mass of Ca₃(PO₄)₂ is:
Formula mass = (3 x 40.08 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)
= 120.24 g/mol + 61.94 g/mol + 128.00 g/mol
= 310.18 g/mol.
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