Mercury metal react beside oxygen to form the bright ginger, mercury(II) oxide...?
Mercury metal reacts with oxygen to form the bright red, mercury(II) oxide
2Hg+O2 --> 2HgO
if 200g of Hg and 30g of O2 are mixed, which is the limiting reagent?
and how many grams of HgO can be formed from this mixture?
A sample of barium chloride beside a mass of 1g was dissolved in river and diluted to 100 mL in a volumetric flask. What is the concentration of the barium chloride?
A 1.23 mL sample of a 1.25M solution of barium chloride be diluted to 25 mL. What is the final concentration of the diluted solution?
What mass of sodium sulfide is required to prepare 250 mL of 0.2M solution?
Answers:
Moles Hg = 200 g / 200.59 g/mol=0.997
moles O2 = 30 g/32 g/mol=0.974
Hg is the limiting reagent ( 0.974 x 2 =1.95 moles of Hg are needed)
moles HgO = 0.997
mass HgO = 0.997 mol x 216.59 g/mol=215.9 g
moles BaCl2 = 1 g/208.23 g/mol=0.00480
[BaCl2] = 0.00480/0.100 L = 0.0480 M
Moles BaCl2 = 0.00123 L x 1.25 M=0.00154
new concentration = 0.00154/ 0.025 L=0.0616 M
Moles = 0.250 L x 0.2 = 0.05
mass = 0.05 x 78 g/mol = 3.9 g
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2Hg+O2 --> 2HgO
if 200g of Hg and 30g of O2 are mixed, which is the limiting reagent?
and how many grams of HgO can be formed from this mixture?
A sample of barium chloride beside a mass of 1g was dissolved in river and diluted to 100 mL in a volumetric flask. What is the concentration of the barium chloride?
A 1.23 mL sample of a 1.25M solution of barium chloride be diluted to 25 mL. What is the final concentration of the diluted solution?
What mass of sodium sulfide is required to prepare 250 mL of 0.2M solution?
Answers:
Moles Hg = 200 g / 200.59 g/mol=0.997
moles O2 = 30 g/32 g/mol=0.974
Hg is the limiting reagent ( 0.974 x 2 =1.95 moles of Hg are needed)
moles HgO = 0.997
mass HgO = 0.997 mol x 216.59 g/mol=215.9 g
moles BaCl2 = 1 g/208.23 g/mol=0.00480
[BaCl2] = 0.00480/0.100 L = 0.0480 M
Moles BaCl2 = 0.00123 L x 1.25 M=0.00154
new concentration = 0.00154/ 0.025 L=0.0616 M
Moles = 0.250 L x 0.2 = 0.05
mass = 0.05 x 78 g/mol = 3.9 g
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