Chemistry Quiz

How many moles are in 235 grams of silicon carbide?

Answer explanation

To determine the number of moles in 235 grams of silicon carbide, divide the given mass by the molar mass of silicon carbide.

235 g SiC × 1 mol SiC/40.1 g SiC = 5.86 mol

How many grams are in 1.07 moles of AgOH?

Answer explanation

To find the number of grams in 1.07 moles of AgOH, we need to use the molar mass of AgOH. By multiplying the molar mass by the number of moles, we can calculate the grams. The correct choice will provide the correct molar mass of AgOH.

1.07 mol AgOH × 124.88 g AgOH/1 mol AgOH = 134 g

How many grams are in 18.1 moles of manganese (III) hydrogen carbonate?

Answer explanation

To find the number of grams in 18.1 moles of manganese (III) hydrogen carbonate, you need to multiply the molar mass of the compound by the number of moles. The molar mass of manganese (III) hydrogen carbonate is calculated by adding the atomic masses of each element in the compound.

18.1 mol Mn(HCO3)3 × [237.99 g Mn(HCO3)3/1 mol Mn(HCO3)3] = 4310 g

How many grams are in 8.24 x 10^24 particles of Cr2O3?

Answer explanation

The molar mass of Cr2O3 is 151.99 g/mol. Therefore, 8.24 x 10^24 particles of Cr2O3 would weigh 1.25 x 10^23 grams.

How many grams are in 8.36 x 10^19 particles of nitrogen gas?

Answer explanation

8.36 × 10 particles N2 × [1 mol N2/ 6.02 × 1023 particles N2] × [28.01 g N2/1 mol N2 =

0.00389 g

How many grams are in 3.86 moles of CrCr2O7?

Answer explanation

To find the number of grams in 3.86 moles of Cr2O7, we need to multiply the molar mass of Cr2O7 by the number of moles. The molar mass of Cr2O7 is 294.18 g/mol. Therefore, the number of grams in 3.86 moles of Cr2O7 is 1135.85 grams.

How many particles are in 1.23 moles of chromium (II) hydroxide?

Answer explanation

1.23 mol Cr(OH)2 × [6.02 × 1023 particles Cr(OH)2/1 mol Cr(OH)2] = 7.4 × 10^23 particles

How many moles are in 72.7 grams of Sc2O3?

Answer explanation

To find the number of moles, divide the given mass by the molar mass of Sc2O3.

0.527 mol

How many particles are in 492 grams of Al2(CO3)3?

Answer explanation

To find the number of particles, use Avogadro's number and the molar mass of Al2(CO3)3. Calculate the moles of Al2(CO3)3, then multiply by Avogadro's number to find the number of particles.

How many particles are in 75.3 grams of manganese (II) nitrite?

Answer explanation

To determine the number of particles in 75.3 grams of manganese (II) nitrite, we need to use Avogadro's number and the molar mass of the compound.

3.08 × 10^23 particles