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What is the number of moles in 32 grams of O2?
There are 1.0001 moles.
100ml of 0.1M NaOH 50ml of 0.2M BaCl2 or 75ml of 0.15M Na3PO4 are strong electrolytes.Which contains the largest number of ions?
To determine which solution contains the largest number of ions, we need to calculate the total moles of ions produced by each solution. NaOH (100ml of 0.1M): 0.1 moles/L × 0.1 L = 0.01 moles of NaOH → produces 0.01 moles Na⁺ and 0.01 moles OH⁻ = 0.02 moles of ions. BaCl2 (50ml of 0.2M): 0.2 moles/L × 0.05 L = 0.01 moles of BaCl2 → produces 0.01 moles Ba²⁺ and 0.02 moles Cl⁻ = 0.03 moles of ions. Na3PO4 (75ml of 0.15M): 0.15 moles/L × 0.075 L = 0.01125 moles of Na3PO4 → produces 0.03375 moles Na⁺ and 0.01125 moles PO4³⁻ = 0.045 moles of ions. Therefore, Na3PO4 contains the largest number of ions.
How many moles of rice grains are equal to 1.807x1024 grains of rice?
To calculate the number of moles of rice grains in 1.807 x 10²⁴ grains, we use Avogadro's number, which is approximately 6.022 x 10²³ grains per mole. Dividing the total number of grains by Avogadro's number gives us: [ \text{Moles of rice} = \frac{1.807 \times 10^{24} \text{ grains}}{6.022 \times 10^{23} \text{ grains/mole}} \approx 3.00 \text{ moles} ] Thus, 1.807 x 10²⁴ grains of rice is approximately 3.00 moles.
How mmany moles of rice grains are equal to 1.807 x 10 grains of rice?
To find the number of moles of rice grains in 1.807 x 10^24 grains, you can use Avogadro's number, which is approximately 6.022 x 10^23 grains per mole. Divide the total number of grains by Avogadro's number: 1.807 x 10^24 grains / 6.022 x 10^23 grains/mole ≈ 3.00 moles of rice grains. Therefore, 1.807 x 10^24 grains of rice is approximately 3.00 moles.
What is the empirical formula for 52.5 percent oxygen 25.4 percent phosphorus 22.1 percent aluminum?
Well first we assume a 100 g sample and then convert the elements into moles. We then divide them all by the number of moles that is the lowest of the 3. This procedure gives us AlPO4
How many moles of Na2CO3 are there in 10.0 L of 2.0 M soluton?
Molarity = moles of solute/Liters of solution Or, for our purposes, Moles of solute = Liters of solution * Molarity Moles Na2CO3 = 10.0 Liters * 2.0 M = 20 moles Na2CO3 --------------------------
How many moles is 0.75g of Na2CO3?
The formula mass of sodium carbonate, Na2CO3 is 2(23.0) + 12.0 + 3(16.0) = 106.0Amount of Na2CO3 = mass of sample/molar mass = 0.75/106.0 = 0.00708mol There are 0.00708 moles of Na2CO3 in a 0.75g pure sample.
How many moles of ions are produced by the dissociation of 0.5 mol of Na2CO3?
When Na2CO3 dissociates, it produces 3 moles of ions: 2 moles of Na+ ions and 1 mole of CO3^2- ions. So, if you have 0.5 moles of Na2CO3, you would produce 1.5 moles of ions in total.
If 1 cup is equal to 251.8 grams how many moles of Na2CO3 are used?
To determine how many moles of Na2CO3 are used, you need to know the molar mass of Na2CO3. It is 105.99 g/mol. Then, you can calculate the number of moles using the given mass: mass in grams / molar mass = moles of Na2CO3.
How many grams of sodium carbonate is needed to react with 18.1 ml of 0.23M HCL?
Balanced equation always first!! Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O Find moles HCl by---Molarity = moles of solute/Liters of solution (18.1 ml = 0.0181 L) 0.23 M HCl = moles HCl/0.0181 liters = 0.004163 moles HCl Drive back against sodium carbonate stoichometrically 0.004163 moles HCl (1 mole Na2CO3/2 mole HCl)(105.99 grams/1 mole Na2CO3) = 0.22 grams Na2CO3 needed --------------------------------------------
How many moles of sodium carbonate are present in 212g?
To calculate the number of moles of sodium carbonate in 212g, you need to first determine the molar mass of Na2CO3 (sodium carbonate). The molar mass of Na2CO3 is 105.99 g/mol. Then, you divide the given mass (212g) by the molar mass to find the number of moles (212g / 105.99 g/mol = 1.999 moles).
How many grams are in 0.577 mol Na2CO3?
To convert moles to grams, you need to multiply the number of moles by the molar mass of the substance. The molar mass of Na2CO3 (sodium carbonate) is 105.99 g/mol. So, to find the grams in 0.577 mol of Na2CO3, you would calculate 0.577 mol * 105.99 g/mol = approximately 60.98 grams of Na2CO3.
Convert the mass of Na2CO3 to moles with the molecular weight of 105.989?
the answer to your question is 0.0004 g/mol.
How many grams of NaOH are formed from reacting 120g sodium carbonate with excess calcium hydroxide?
Write out the equation, and remember to balance each side.Na2CO3 + Ca(OH)2 --> 2NaOH + CaCO3Molecular WeightsNa2CO3: 106 grams/moleNaOH: 40 grams/moleAlways convert your reagents into moles.(120g Na2CO3) x (1 mole Na2CO3/106 grams Na2CO3) = 1.132 molesAccording to the balanced equation, 1 molecule of Na2CO3 generates 2 molecules of NaOH.(1.132 moles Na2CO3) x (2 moles NaOH/1 mole Na2CO3) = 2.264 moles NaOHNow determine the number of grams from 2.264 moles of NaOH.(2.264 moles NaOH) x (40 grams/ 1 mole NaOH) = 90.57 grams NaOH formed.To prevent rounding off too many times, carry out the dimensional analysis in one step:(120g Na2CO3) x (1 mole Na2CO3/106 grams Na2CO3) x(2 moles NaOH/1 mole Na2CO3) x (40 grams/ 1 mole NaOH) = 90.57 grams NaOH
How many grams of HCL needed to react with 4000 g of Na2co3?
The balanced chemical equation for this reaction is 2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2. From the equation, 1 mole of Na2CO3 reacts with 2 moles of HCl. Calculate the number of moles of Na2CO3 in 4000g, then use the mole ratio to find the moles of HCl needed. Finally, convert moles of HCl to grams.
How many moles of washing soda are there in 5g?
Washing soda is sodium carbonate, Na2CO3. Using the atomic weights from the periodic table and the subscripts in the formula, the molar mass of Na2CO3 = 106g/mol. 5g Na2CO3 x (1mol Na2CO3/106g/mol) = 0.05mol Na2CO3
How are 0.50 mol Na2CO3 and 0.50M Na2CO3 different?
0.50 mol of Na2CO3 represents a fixed quantity of the compound (50% of a mole), whereas 0.50M Na2CO3 indicates the concentration of Na2CO3 in a solution (0.50 moles per liter). The former is a measure of the amount of the substance, while the latter is a measure of its concentration in solution.
