CHEM 1341 – Practice ACS Exam

How much of a 6.00M HNO₃ stock solution is needed to prepare 400. mL of 0.0250M HNO₃ solution?

• 24.0 mL

• 16.7 mL

• 1.67 mL

• 2.40 mL

Which ground-state atom has an electron configuration described by the following orbital diagram?

• phosphorus

• nitrogen

• sulfur

• carbon

Tags

NGSS.HS-PS1-1

Which quantum number primarily determines the average distance of an electron from the nucleus?

• Principal quantum number

• Angular momentum quantum number

• Magnetic quantum number

• Spin quantum number

Which molecule corresponds to a planar geometry?

• NH3

• H3O+

Of the following four sets of quantum numbers, which set is not allowed?

• n = 3, l = 2, m_l = -1, m_s = -1/2

• n = 4, l = 2, m_l = -2, m_s = -1/2

• n = 5, l = 2, m_l = +2, m_s = +1/2

• n = 5, l = 2, m_l = +3, m_s = +1/2

Which of the following illustrates a physical process occurring?

• a white precipitate forms when solutions of silver nitrate and sodium chloride are mixed.

• A piece of magnesium metal reacts with hydrochloric acid, producing bubbles

• Solid iodine sublimes to form a violet gas when gently heated.

• A strip of copper turns green after prolonged exposure to moist air

Tags

NGSS.MS-PS1-4

Of the molecules listed below (a – d), which one has a net dipole moment?

• NH3

• AlCl3

• XeF4

• SF6

Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions. A mixture of 82.49 g of aluminum and 117.65 g of oxygen is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

• Oxygen is the limiting reactant; 19.81 g of aluminum remain.

• Oxygen is the limiting reactant; 35.16 g of aluminum remain.

• Aluminum is the limiting reactant; 16.70 g of oxygen remain.

• Aluminum is the limiting reactant; 35.16 g of oxygen remain.

• Aluminum is the limiting reactant; 44.24 g of oxygen remain.

Tags

NGSS.HS-PS1-7

• 711 nm, 1.87 eV

• 663 nm, 1.87 eV

Tags

NGSS.HS-PS4-1

Aluminum nitrate (Al(NO₃)₃) reacts with sodium sulfide (Na₂S) to form a precipitate: 2Al(NO₃)₃(aq) + 3Na₂S(aq) → Al₂S₃(s) + 6NaNO₃(aq) When reacting 35.0 mL of the aluminum nitrate solution with an excess of sodium sulfide, 1.420 g of the aluminum sulfide precipitate is recovered. What was the molar concentration of the aluminum nitrate solution?

• 0.382 M

• 0.407 M

• 0.433 M

• 0.540 M

Tags

NGSS.HS-PS1-7

NGSS.HS-PS1-5