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How many grams of AgNO3 are required to make 500.0 mL of a 0.10 M solution?
Ah, what a lovely question! To make a 0.10 M solution of AgNO3 in 500.0 mL, we can use the formula: moles = molarity x volume (in liters). First, convert 500.0 mL to liters by dividing by 1000. Then, multiply the molarity (0.10 M) by the volume in liters to find the moles of AgNO3 needed. Finally, convert moles to grams using the molar mass of AgNO3. Happy calculating!
How many moles are in 680g of AgNO3?
Roughly 4 moles.
Determine the number of formula units that are in 0.688 moles of AgNO3?
0.688 moles*6.02x1023=4.14x1023 Formula units
How many silver atoms would be in 4.55 moles of AgNO3?
To find the number of silver atoms in 4.55 moles of AgNO3, first calculate the molar mass of AgNO3 which is 169.87 g/mol. Then set up a ratio using Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. The calculation would be 4.55 moles x (6.022 x 10^23 atoms/mol) = 2.74 x 10^24 silver atoms in 4.55 moles of AgNO3.
How many moles of Ag are produced when starting with 6.2 moles of AgNO3?
6,2 moles of silver
What is the molarity of a solution if 255 grams AgNO3 is dissolved in 1500 mL of solution?
Get moles silver nitrate. 255 grams AgNO3 (1 mole AgNO3/169.91 grams) = 1.5008 moles AgCO3 --------------------------------Now; Molarity = moles of solute/Liters of solution ( 1500 ml = 1.5 Liters ) Molarity = 1.5008 moles AgNO3/1.5 Liters = 1.00 M AgNO3 ---------------------
How many grams of AgNO3 are required to make 500.0 mL of a 0.10 M solution?
Ah, what a lovely question! To make a 0.10 M solution of AgNO3 in 500.0 mL, we can use the formula: moles = molarity x volume (in liters). First, convert 500.0 mL to liters by dividing by 1000. Then, multiply the molarity (0.10 M) by the volume in liters to find the moles of AgNO3 needed. Finally, convert moles to grams using the molar mass of AgNO3. Happy calculating!
How many moles of AgNO3 does 85 grams of AgNO3 represents?
To find the number of moles, you need to divide the given mass (85 grams) by the molar mass of AgNO3 (169.87 g/mol). 85 grams of AgNO3 represents 0.500 moles.
How many moles of agno3 are needed to prepare 0.50 l of a 4.0 m solution?
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
How many moles are in 680g of AgNO3?
Roughly 4 moles.
How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?
Balanced equation first! AgNO3 + NaCl -> AgCl + NaNO3 all one to one, get moles AgNO3 3.82 moles NaCl (1 mole AgNO3/1 mole NaCl) = 3.82 moles AgNO3 ------------------------------- Molarity = moles of solute/Liters of solution 0.117 M AgNO3 = 3.82 moles AgNO3/Liters Liters = 3.82/0.117 = 32.6 Liters which is 32600 milliliters which is unreasonable; check answer if you can
Determine the number of formula units that are in 0.688 moles of AgNO3?
0.688 moles*6.02x1023=4.14x1023 Formula units
How many silver atoms would be in 4.55 moles of AgNO3?
To find the number of silver atoms in 4.55 moles of AgNO3, first calculate the molar mass of AgNO3 which is 169.87 g/mol. Then set up a ratio using Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. The calculation would be 4.55 moles x (6.022 x 10^23 atoms/mol) = 2.74 x 10^24 silver atoms in 4.55 moles of AgNO3.
How many moles of Ag are produced when starting with 6.2 moles of AgNO3?
6,2 moles of silver
How many moles of silver are present in 32.46g of AgNO3?
To find the number of moles of silver in 32.46g of AgNO3, first calculate the molar mass of AgNO3 (169.87 g/mol). Then, divide the given mass by the molar mass to find the number of moles (32.46g / 169.87 g/mol ≈ 0.191 moles). Since there is one mole of Ag in one mole of AgNO3, there are 0.191 moles of silver present.
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
How many moles of AgCl will be produced from 83.0 g of AgNO3 assuming NaCl is available in excess?
The balanced chemical equation for this reaction is: AgNO3 + NaCl -> AgCl + NaNO3 From this equation, we can see that 1 mole of AgNO3 produces 1 mole of AgCl. Since the molar mass of AgNO3 is 169.87 g/mol, 83.0 g of AgNO3 is equivalent to 0.488 moles. Therefore, 0.488 moles of AgCl will be produced.
