Concept of Mole and Chemical Counting

What is the unit used to measure the amount of any chemical substance?

The Avogadro number is—

One mole of any substance contains how many particles?

The mass of 1 mole of carbon atoms is—

The mass of 1 mole $\mathrm{H_2O}$ is—

The mass of 1 mole NaCl is—

12 g of carbon contains how many C atoms?

The number $6.023 \times 10^{23}$ is named after—

Identification of an element and its properties is—

Determining the amount of a substance is called—

1 mole of an ion contains—

The mass of 1 mole of H atoms is—

The molecular mass of $\mathrm{Cl_2}$ is—

1 mole of $\mathrm{Cl_2}$ equals—

The molecular mass of $\mathrm{H_2SO_4}$ is—

The molecular mass of $\mathrm{Na_2CO_3}$ is—

The molecular mass of $\mathrm{CuSO_4 \cdot 5H_2O}$ is—

The molecular mass of NaCl is—

The molecular mass of $\mathrm{NH_3}$ is—

The molecular mass of $\mathrm{CO_2}$ is—

The molecular mass of $\mathrm{CaCO_3}$ is—

The molecular mass of $\mathrm{CH_4}$ is—

The molecular mass of $\mathrm{Al_2(SO_4)_3}$ is—

The molecular mass of $\mathrm{MgO}$ is—

The molecular mass of $\mathrm{CaCl_2}$ is—

The molecular mass of NaOH is—

The molecular mass of HCl is—

STP refers to—

The molar volume of a gas at STP is—

Volume of 2 g hydrogen gas at STP is—

Volume of 5 mole $\mathrm{CO_2}$ at STP is—

Number of molecules in 1 L of $\mathrm{CO_2}$ at STP is approx—

Volume of 10 g $\mathrm{H_2}$ gas at STP is—

The relation $n = V/22.4$ applies to—

Volume of 0.5 mole $\mathrm{O_2}$ at STP—

Volume of 4 g $\mathrm{O_2}$ at STP—

11.2 L of $\mathrm{O_2}$ at STP equals how many moles?

112 L of $\mathrm{N_2}$ at STP equals how many moles?

Volume of 2 mole $\mathrm{O_2}$ at STP—

22.4 L $\mathrm{CO_2}$ at STP contains molecules—

At STP, 1.12 L $\mathrm{O_2}$ contains molecules—

In the formula $n = w/M$, what does $n$ represent?

Approximate mass of 1 $\mathrm{H_2O}$ molecule is—

Number of moles of $\mathrm{H_2O}$ in 5 g $\mathrm{H_2O}$—

The mass of $2 \times 10^{23}$ molecules of $\mathrm{CO_2}$ is approx—

Number of molecules in 10 g $\mathrm{CaCO_3}$—

Number of $\mathrm{H_2SO_4}$ molecules in 1 g $\mathrm{H_2SO_4}$—

Number of S atoms in 1 mole $\mathrm{H_2SO_4}$—

Number of O atoms in 1 mole $\mathrm{H_2SO_4}$—

Number of $\mathrm{Na^+}$ ions in 1 mole NaCl—

Mass of 1 mole Cu atoms—

98 g $\mathrm{H_2SO_4}$ contains how many moles?

44 g $\mathrm{CO_2}$ contains how many moles?

32 g $\mathrm{O_2}$ contains how many molecules?

18 g $\mathrm{H_2O}$ contains how many H atoms?

58.5 g NaCl equals how many moles?

Number of H atoms in 1 mole $\mathrm{H_2}$—

Number of O atoms in 2 mole $\mathrm{O_2}$—

Solution = ?

The substance dissolved in a solvent is called—

The solvent in salt water is—

A solution with comparatively less solute is called—

A solution with comparatively more solute is called—

When 1 mole solute is dissolved in 1 L solution at fixed temperature, it is called—

The molarity of a semimolar solution is—

The molarity of a decimolar solution is—

Relation between molarity (S), volume (V in mL), mass (w) and molecular mass (M)—

NaCl required to prepare 250 mL 0.2 M NaCl solution—

Mass of $\mathrm{Na_2CO_3}$ in 2 L 0.1 M $\mathrm{Na_2CO_3}$ solution—

Molarity of solution with 20 g $\mathrm{Na_2CO_3}$ in 250 mL—

Volume containing 20 g $\mathrm{Na_2CO_3}$ at 0.75 M—

$\mathrm{Na_2CO_3}$ needed for 200 mL semimolar solution—

Molarity of 4 g NaOH in 100 mL solution—

Approximate molarity of 4 g HCl in 100 mL solution—

$\mathrm{Na_2CO_3}$ needed for 250 mL 0.1 M solution—

When the solvent is water, the solution is called—

The number of moles of solute per liter of solution is—

NaCl required for 1 L of 1 M NaCl solution—

$\mathrm{H_2SO_4}$ needed for 100 mL 0.1 M $\mathrm{H_2SO_4}$—

Mass of $\mathrm{H_2SO_4}$ in 200 mL 0.25 M $\mathrm{H_2SO_4}$—

$\mathrm{Na_2CO_3}$ needed for 1 L 0.1 M solution—

Percent composition of an element in a compound equals—

Percent composition of H in HCl—

Percent composition of Cl in HCl—

Percent composition of H in $\mathrm{H_2O}$—

Percent composition of O in $\mathrm{H_2O}$—

Percent composition of H in $\mathrm{H_2SO_4}$—

Percent composition of S in $\mathrm{H_2SO_4}$—

Percent composition of O in $\mathrm{H_2SO_4}$—

Percent composition of Al in $\mathrm{Al_2(SO_4)_3}$—

Percent composition of S in $\mathrm{Al_2(SO_4)_3}$—

Percent composition of O in $\mathrm{Al_2(SO_4)_3}$—

Percent composition of C in $\mathrm{CaCO_3}$—

Percent composition of Ca in $\mathrm{CaCO_3}$—

Percent composition of O in $\mathrm{CaCO_3}$—

Percent composition of N in $\mathrm{NH_3}$—

Percent composition of C in $\mathrm{CH_4}$—

The formula expressing the simplest atom ratio of a molecule is—

The empirical formula of $\mathrm{H_2O_2}$ is—

The molecular formula of glucose is—

The empirical formula of glucose ($\mathrm{C_6H_{12}O_6}$) is—

The empirical formula of benzene ($\mathrm{C_6H_6}$) is—

The empirical formula of ethene ($\mathrm{C_2H_4}$) is—

In molecular formula = (empirical formula)$_n$, $n$ equals—

C = 92.31%, H = 7.69%, molecular mass = 78. Molecular formula is—

H = 11.11%, O = 88.89%. Empirical formula is—

H = 2.04%, S = 32.65%, O = 65.30%. Empirical formula is—

C = 40%, H = 6.67%, O = 53.33%. Empirical formula is—

The empirical formula of $\mathrm{H_2SO_4}$ is—

The molecular formula of water is $\mathrm{H_2O}$, its empirical formula is—

A compound contains 3 g of carbon and 8 g of oxygen. Empirical formula is—

The substances that begin a chemical reaction are called—

The new substances formed at the end of a reaction are—

In a chemical equation, reactants are written on—

Products are written on—

The state symbol for gas is—

The state symbol for solid is—

The state symbol for liquid is—

The state symbol for an aqueous solution is—

Heat applied to a reaction is shown above the arrow as—

The basis for balancing chemical equations is—

In $\mathrm{CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O}$, the state of $\mathrm{CO_2}$ is—

Balanced equation of $\mathrm{Mg + HCl \to MgCl_2 + H_2}$ is—

Balanced equation of $\mathrm{Na_2CO_3 + HCl \to NaCl + H_2O + CO_2}$ is—

Balanced equation of $\mathrm{H_2 + O_2 \to H_2O}$ is—

Balanced equation of $\mathrm{Al_2O_3 + HCl \to AlCl_3 + H_2O}$ is—

In $\mathrm{2Mg(NO_3)_2 \xrightarrow{\Delta} 2MgO + ? + O_2}$, the missing product is—

The reaction of solid C with $\mathrm{O_2}$ produces—

In $\mathrm{CaCO_3 + 2HCl \to CaCl_2 + ? + H_2O}$, the missing product is—

The branch of chemistry that deals with calculations of reactant and product amounts is—

In $\mathrm{2Mg + O_2 \to 2MgO}$, $\mathrm{O_2}$ needed for 5 g Mg is approximately—

In $\mathrm{2Mg + O_2 \to 2MgO}$, MgO produced from 2 g Mg is approximately—

In $\mathrm{N_2 + 3H_2 \to 2NH_3}$, number of $\mathrm{NH_3}$ molecules from 5 $\mathrm{N_2}$ molecules—

Moles of $\mathrm{O_2}$ required to produce 6 moles of water—

In $\mathrm{2Mg + O_2 \to 2MgO}$, $\mathrm{O_2}$ needed to produce 10 g MgO is—

In $\mathrm{CaCO_3 \to CaO + CO_2}$, $\mathrm{CO_2}$ produced from 100 g $\mathrm{CaCO_3}$—

In $\mathrm{CaCO_3 \to CaO + CO_2}$, CaO produced from 100 g $\mathrm{CaCO_3}$—

In $\mathrm{2Mg + O_2 \to 2MgO}$, moles of $\mathrm{O_2}$ for 2 mole MgO—

In $\mathrm{N_2 + 3H_2 \to 2NH_3}$ at STP, $\mathrm{NH_3}$ volume from 4 L $\mathrm{N_2}$—

In $\mathrm{C + O_2 \to CO_2}$, $\mathrm{O_2}$ needed for 12 g C is—

In $\mathrm{N_2 + 3H_2 \to 2NH_3}$, $\mathrm{NH_3}$ produced from 28 g $\mathrm{N_2}$—

Stoichiometric calculations require—

In $\mathrm{2Mg + O_2 \to 2MgO}$, Mg required to produce 80 g MgO—

Among multiple reactants, the one present in less than the stoichiometric amount is—

The amount of product formed is determined by—

With 5 g $\mathrm{H_2}$ and 75 g $\mathrm{Cl_2}$, the limiting reactant is—

In the above reaction, the leftover is approximately—

With 4 Mg atoms and 4 $\mathrm{O_2}$ molecules in $\mathrm{2Mg + O_2 \to 2MgO}$, the limiting reactant is—

With 70 Mg atoms and 30 $\mathrm{O_2}$ molecules, the limiting reactant is—

The number of Cl atoms in 75 g $\mathrm{Cl_2}$ is approximately—