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What is the density of an object if its volume is 40cm3 and its mass is 120grams?
Wiki User
∙ 9y ago
Density = Mass/Volume so the density is 120/40 grams per cm3 = 3 grams per cubic cm.
Wiki User
∙ 9y ago
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What is the density of a substances with a volume of 40cm3 and mass of 62.8g?
Density = Mass/Volume = 62.8/40 = 1.57 g/cm3.
What is the density of a material that has a volume of 40 cm cubed and a mass of 2 grams?
density = mass/volume = 2g/40cm3 = 0.05g/cm3
What is the volume of 40cm3?
40 cubic centimeters.
What is the density of a substances with a volume of 40cm3 and mass of 62.8g?
Density = Mass/Volume = 62.8/40 = 1.57 g/cm3.
What is the density of a material that has a volume of 40 cm cubed and a mass of 2 grams?
density = mass/volume = 2g/40cm3 = 0.05g/cm3
What is the mass of a piece of pyrite that has a volume of 40cm3?
Iron pyrite (FeS2) has a density of 4.8 to 5.0 grams per cm3. A piece of volume 40 cm3 would, therefore have a mass of volume*density = 192 to 200 grams.
What is the volume of 40cm3?
40 cubic centimeters.
40cm3 how many oz?
40cm3 equals 1.35 fluid ounces.
What is the volume of a solid that is 4cm height 5 cm length and width 2 cm with an answer cm cubed?
4 x 5 x 2 = 40cm3
What is the volume of a rectangular prism with a length of 4 cm a width of 2 cm and a height of 5 cm?
A rectangular prism with a length of 4 cm, a width of 2 cm and a height of 5 cm has a volume of 40cm3
What is the volume of a lead ball with a mass of 454 g?
Easy. Density is equal to mass / volume (that's why you have a mass and volume unit g/cm3) So 11.34g/cm3 = 454g / volume and with basic algebra: 454g / 11.34g/cm3 = volume volume ~ 40.04cm3 or 40,035273368606701940035273368607cm3
What volume of hydrogen chloride gas is produce by the reaction of 40cm3 of chlorine with hydrogen?
Chlorine exists in the form Cl2. When it reacts with hydrogen, the reaction is Cl2 + H2 -> 2HCl. Given forty cubic centimeters of chlorine (which has a density of .003214 g/cm3 and a molar mass of 70.906 g/mol) this equates to 1.81 millimoles of chlorine. Each mole of chlorine produces 2 of hydrogen chloride, so this equates to 3.62 millimoles of HCl. HCl has a molar mass 36.461 g/mol and a density of 1.18 g/cm3. This gives .132 grams and .112 cubic centimeters of HCl as the final product.
What volume of Hydrogen chloride gas is produced by 40cm3 of chlorine with hydrogen?
It is necessary first to determine how many moles of chlorine there are in the volume specified. Using the ideal gas law, 40 cm3 of chlorine is equivalent to 0.129 grams of Cl2, or 3.63 millimoles. Next, the stoichiometric equation is set up. X H2 + Y Cl2 -- reacts --> Z HCl. It can be clearly seen that X = 1, Y = 1, and Z = 2. This means that for every mole of chlorine that reacts, 2 moles of hydrogen chloride is produced. Then, we have 7.26 millimoles of HCl. Reversing the ideal gas law, this means the resulting volume of HCl gas is .224 cm3. The reason for the smaller volume is due to the differences in the molar weight of chlorine and hydrogen chloride (70.906 g/mol and 36.461 g/mol respectively).
What is the mass of a 10.0 cubic centimeters of lead?
Specific Gravity of lead is 11.35. Books state that weight=(sp.gr.)Xvolume. So weight of 40cm3 block of lead = 11.35 x 40 gms = 254 gms I understand this is wieght. Do we need to devide this weight by accelaration due gravity to calculate mass ? If whole block is converted to Energy as per Einstein Equation E=254 x c x c or E=254 x c x c / g where c = velocity of light g = accelaration due gravity I had asked similar question earlier. I could not get satisfactory answer. Do somebody have answer?
What is the concentration of excess H2SO4when50cm3 of 0.200moldm-3 NaOH is added to 40cm3 of 0.250 moldm-3 H2SO4.?
moles H2SO4 = 0.040 L x 0.250 mol/L = 0.01 moles H2SO4moles NaOH = 0.05 L x 0.200 mol/L = 0.01 moles NaOHReaction: 2NaOH + H2SO4 = Na2SO4 + 2H2OIt takes 2 moles NaOH for each 1 mole H2SO4Moles H2SO4 used up by 0.01 moles NaOH = 0.005 molesmoles H2SO4 remaining = 0.005 moles H2SO4Total volume = 50 cm3 + 40 cm3 = 90 cm3 = 0.090 LFinal [H2SO4] = 0.005 moles/0.090 L = 0.055 M (to 2 significant figures)
