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2 years ago

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B. Steve selects an amino acid that you wrote for part A and dissolves 0.1 moles in 1 liter of water. He adjusts the pH to 7.2,

but then absentmindedly adds 0.03 more moles of HCl to the solution. Is the solution still a good buffer

1 answer:

Lady_Fox [76]2 years ago

8 0

Answer:

i asked to where my uncle shoes he told my no

Explanation:

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If 5.61 g of ammonium nitrate (NH4NO3) is dissolved in 155 mL of water to make an instant cold pack, what is the molarity of the

kifflom [539]

Answer:

0.452 M

Explanation:

Remember that to get molarity, you have to go thru an equation:

M=mol/V (in liters)

So...

5.61 g NH4NO3 x 1 mol NH4NO3/80.043 g

=0.0701 mol NH4NO3

Convert mL to L

=155--->0.155 L

Plug in and divide.

0.0701/0.155

=0.452 M

7 0

2 years ago

14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its

masha68 [24]

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

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2 years ago

1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?

Lyrx [107]

Answer: Option (b) is the correct answer.

Explanation:

The energy necessary to remove an electron from a gaseous atom or ion is known as ionization energy.

This means that smaller is the size of an atom more amount of energy has to be supplied to it in order to remove the valence electron. This is because in small atom or element there will be strong force of attraction between the nucleus and electrons.

So, high amount of energy has to be supplied to remove the valence electrons.

As electronic configuration of helium is $1s^{2}$ . So, due to completely filled valence shell it is more stable in nature.

As a result, we need to provide very high amount of energy to remove an electron from a helium atom.

Thus, we can conclude that out of the given options helium element would the first ionization energy of the atom be higher than that of the diatomic molecule.

7 0

2 years ago

A gas occupies 22.4 l at stp and 14.5 l at 100c and 2.00 atm pressure. how many moles of gas did the system gain or lose?

azamat

<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>

6 0

2 years ago

You have two 500.0 ml aqueous solutions. solution a is a solution of a metal nitrate that is 8.246% nitrogen by mass the ionic c

almond37 [142]

1) Answer is: the ionic compound in the solution b is K₂CrO₄ (potassium chromate).

Ionic compound in solution b has two potassiums (oxidation number +1), one chromium (oxidation number +6) and four oxygens. Oxidation number of oxygen is -2 and compound has neutral charge:

2 · (+1) + 6 + x · (-2) = 0.

x = 4; number of oxygen atoms.

2) Answer is: the ionic compound in solution a is AgNO₃ (silver nitrate).

ω(N) = 8.246% ÷ 100%.

ω(N) = 0.08246; mass percentage of nitrogen.

M(MNO₃) = M(N) ÷ ω(N).

M(MNO₃) = 14 g/mol ÷ 0.08246.

M(MNO₃) = 169.8 g/mol; molar mass of metal nitrate.

M(M) = M(MNO₃) - M(N) - 3 · M(O).

M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol.

M(M) = 107.8 g/mol; atomic mass of metal, this metal is silver (Ag).

3) Balanced chemical reaction:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).

Ionic reaction:

2Ag⁺(aq) + 2NO₃(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq).

Net ionic reaction: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s).

Answer is: the blood-red precipitate is silver chromate (Ag₂CrO₄).

4) m(Ag₂CrO₄) = 331.8 g; mass of solid silver chromate.

n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄).

n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol.

n(Ag₂CrO₄) = 1 mol; amount of silver chromate.

From balanced chemical reaction: n(Ag₂CrO₄) : n(AgNO₃) = 1 : 2.

n(AgNO₃) = 2 · 1 mol.

n(AgNO₃) = 2 mol.

m(AgNO₃) = n(AgNO₃) · M(AgNO₃).

m(AgNO₃) = 2 mol · 169.8 g/mol.

m(AgNO₃) = 339.6 g; mass of silver nitrate.

m(AgNO₃) = m(K₂CrO₄).

m(K₂CrO₄) = 339.6 g; mass of potassium chromate.

n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄).

n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol.

n(K₂CrO₄) = 1.75 mol; amount of potassium chromate.

5) Chemical reaction of dissociation of silver nitrate in water:

AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq).

V(solution a) = 500 mL ÷ 1000 mL/L.

V(solution a) = 0.5 L; volume of solution a.

c(AgNO₃) = n(AgNO₃) ÷ V(solution a).

c(AgNO₃) = 2 mol ÷ 0.5 L.

c(AgNO₃) = 4 mol/L = 4 M.

From dissociation of silver nitrate: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻).

c(Ag⁺) = 4 M; the concentration of silver ions in the original solution a.

c(NO₃⁻) = 4 M; the concentration of silver ions in the original solution a.

6) Chemical reaction of dissociation of potssium chromate in water:

K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq).

V(solution b) = 500 mL ÷ 1000 mL/L.

V(solution b) = 0.5 L; volume of solution b.

c(K₂CrO₄) = n(K₂CrO₄) ÷ V(solution b).

c(AgNO₃) = 1.75 mol ÷ 0.5 L.

c(AgNO₃) = 3.5 mol/L = 3.5 M.

From dissociation of silver nitrate: c(K₂CrO₄) = c/2(K⁺) = c(CrO₄²⁻).

c(K⁺) = 7 M; the concentration of potassium ions in the original solution b.

c(CrO₄²⁻) = 3.5 M; the concentration of silver ions in the original solution b.

7) V(final solution) = V(solution a) + V(solution b).

V(final solution) = 500.0 mL + 500.0 mL.

V(final solution) = 1000 mL ÷ 1000 mL/L.

V(final solution) = 1 L.

n(NO₃⁻) = 2 mol.

c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution)

c(NO₃⁻) = 2 mol ÷ 1 L.

c(NO₃⁻) = 2 M; the concentration of nitrate anions in final solution.

8) in the solution b there were 3.5 mol of potassium cations, but one part of them reacts with 2 moles of nitrate anions:

K⁺(aq) + NO₃⁻(aq) → KNO₃(aq).

From chemical reaction: n(K⁺) : n(NO₃⁻) = 1 : 1.

Δn(K⁺) = 3.5 mol - 2 mol.

Δn(K⁺) = 1.5 mol; amount of potassium anions left in final solution.

c(K⁺) = Δn(K⁺) ÷ V(final solution).

c(K⁺) = 1.5 mol ÷ 1 L.

c(K⁺) = 1.5 M; the concentration of potassium cations in final solution.

4 0

2 years ago