For this you need the Atomic Mass of Ca. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.
125 grams Ca / (40.1 grams) = 3.12 moles Ca
Well, Ca has an atomic mass of 40, so one mole of Ca (6.022x1023 atoms) equals 40g.To get 5kg of Ca, you would times the 40g (one mole) by 125.5kg of Ca has 125x(6.022x1023), or602200000000000000000000 atoms.
To answer this we must first find the molar mass of calcium carbonate. CaCO3Ca= 40.08gC=12.01gO= 16.00g (we have three oxygens so 16.00x3 is 48.00g)40.08+12.01+48.00= 100.09 gNow that we have the molar mass we can find how many grams there are:1.25 moles CaCo3 x (100.09 g CaCO3/ 1 mole CaCO3)= 125.11 grams CaCO3Therefore we'd have about 125 grams of CaCO3
This question is solved with the help of mole concept . 1 mole of anhydrous calcium carbonate weighs 40+12+48=100 gm . 1.25 mole of similar anhydrous calcium carbonate will be 100* 1.25 = 125 gm
The Molarity of the solution is .901
There are at least three kinds of sodium sulfide, but assuming that the question refers to the most common one with the formula Na2S, its gram formula mass, the mass corresponding to molar mass for covalently bonded compounds, is 78.04. Therefore 125.00 constitutes 125.00/78.04 or 1.6017 "moles".
Well, Ca has an atomic mass of 40, so one mole of Ca (6.022x1023 atoms) equals 40g.To get 5kg of Ca, you would times the 40g (one mole) by 125.5kg of Ca has 125x(6.022x1023), or602200000000000000000000 atoms.
To answer this we must first find the molar mass of calcium carbonate. CaCO3Ca= 40.08gC=12.01gO= 16.00g (we have three oxygens so 16.00x3 is 48.00g)40.08+12.01+48.00= 100.09 gNow that we have the molar mass we can find how many grams there are:1.25 moles CaCo3 x (100.09 g CaCO3/ 1 mole CaCO3)= 125.11 grams CaCO3Therefore we'd have about 125 grams of CaCO3
To calculate the amount of KCl needed, we first need to find the number of moles of KCl required using the formula: moles = Molarity x Volume (in L). Then, we convert moles to grams using the molar mass of KCl, which is 74.55 g/mol. Finally, we use the formula: grams = moles x molar mass to find that approximately 6.33 grams of KCl are needed to prepare 125 mL of a 0.720 M solution.
To produce 1 mole of urea, 1 mole of carbon dioxide is needed. The molar mass of urea is 60 grams/mol, and the molar mass of carbon dioxide is 44 grams/mol. Therefore, to produce 125 grams of urea, 125 grams/60 grams/mol = 2.08 moles of urea is needed. This means 2.08 moles of carbon dioxide is needed, which is 2.08 moles * 44 grams/mol = 91.52 grams of carbon dioxide needed.
This question is solved with the help of mole concept . 1 mole of anhydrous calcium carbonate weighs 40+12+48=100 gm . 1.25 mole of similar anhydrous calcium carbonate will be 100* 1.25 = 125 gm
1000 grams= 1 kilogram 125 grams= .125 kilo grams and instead of asking people online, next time DIY!!!!!!!!!!!!!
No such thing as HSO in chemistry. If you're referring to H2SO4, which is sulfuric acid, then 125 grams of it would be: H2SO4 = 98g/mol; 98/1=125/x; solve for x to get about 1.28 moles.
To convert 125 lb to kilograms, you would first convert pounds to ounces (125 lb * 16 oz/lb = 2000 oz) and then convert ounces to grams (2000 oz * 28 g/oz = 56000 g). Finally, convert grams to kilograms by dividing by 1000 (56000 g / 1000 = 56 kg). So, 125 lb is equal to 56 kilograms.
There is 125 milliliters in 125 grams. This is when converting liquid ingredients. To convert grams to milliliters the density of the ingredients needs to be known.
There is 125 grams in a cup. So 450 is 3.6 cups
First, you must find the amount of moles of NaOH, using the concentration and volume given. By lowercase m, I'm assuming you mean molality, or molals of solution, which is the equation:molality (m) = (moles of solute) / (total volume of solution (in liters))To solve for moles of NaOH, your solute, rearrange the equation by multiplying volume on both sides to get:moles solute = (molality)(total volume of solution)Next, just plug in the information you know, which is 500 mL for the total volume and 125 m for the molality.***Volume for concentration problems must be converted to liters, so remember that 1 L = 1000 mLmoles NaOH = (125 m)(0.500 L) = 62.5 molesFinally, convert this to grams by finding the molar mass of NaOH using the periodic table:22.99 + 16.00 + 1.008 = 39.998 g/mol62.5 moles (39.998 g) / (1 mol) =249.875 grams NaOH
Just 125 kg of CaCO3 contains 125000 grams of the substance. I kilogram of any substance is equivalent to 1000 grams of it.