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How many moles in 4.50 grams of silver nitrate?
To find the number of moles in 4.50 grams of silver nitrate (AgNO3), you first need to calculate the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. Then, use the formula: moles = mass/molar mass. So, 4.50 grams of AgNO3 is equal to 0.0265 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 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
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.
4.02 grams of AgNO3 react with excess NaCl how many grams of silver nitrate are produced?
AgNo3(aq) + NaCl(aq) -------------->AgCl (ppt) + No3(-) + Na(+) no silver nitrates are produced it is all consumed. only silver chloride is produced and precipitate . free nitrate and free sodium ions are produced but do not react with each other 1 Mole AgNo3 ------->169.9 gm 1 Mole Nacl ------->58.4 gm 1 Mole AgCl ------->143.3 gm 4.02 gm AgNO3 = (4.02 / 169.9) = 0.02366 M AgCl produced = 0.02366 M = 3.39 gm
How many moles in 4.50 grams of silver nitrate?
To find the number of moles in 4.50 grams of silver nitrate (AgNO3), you first need to calculate the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. Then, use the formula: moles = mass/molar mass. So, 4.50 grams of AgNO3 is equal to 0.0265 moles.
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.
How many moles of Ag are produced when starting with 6.2 moles of AgNO3?
6,2 moles of silver
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 of silver of ions are presented in 32.46 g of AgNO3?
The number of moles is 0,19.
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
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.
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.
How many molecules are there in 65 g of silver nitrate (AgNO3)?
Well, because you have 65g of AgNO3, you have .3826 moles of silver nitrate. This is found by dividing the number of grams you have by the molar mass of silver nitrate (169.9g/mol). Once you know how many moles there are you can then multiply by Avogodro's number (6.022x1023) to obtain the number of molecules. In this case it is 2.304x1023 molecules.
4.02 grams of AgNO3 react with excess NaCl how many grams of silver nitrate are produced?
AgNo3(aq) + NaCl(aq) -------------->AgCl (ppt) + No3(-) + Na(+) no silver nitrates are produced it is all consumed. only silver chloride is produced and precipitate . free nitrate and free sodium ions are produced but do not react with each other 1 Mole AgNo3 ------->169.9 gm 1 Mole Nacl ------->58.4 gm 1 Mole AgCl ------->143.3 gm 4.02 gm AgNO3 = (4.02 / 169.9) = 0.02366 M AgCl produced = 0.02366 M = 3.39 gm
What is 16 grams of oxygen how many moles?
16 grams of oxygen how many moles is 0,5 moles.
