119 grams of uranium is equivalent to 0,5 moles.
To calculate the grams of potassium bromite in 0.280 moles, you would multiply the number of moles by the molar mass of KBrO2, which is 119 g/mol. Therefore, 0.280 moles of KBrO2 would be 0.280 moles * 119 g/mol = 33.32 grams.
Easy...the periodic table gives the average mole weight of each element averaged out over all it's naturally occurring, if any, isotopes, i.e., K is 39.09 Br is 79.9 Add them up and it's (to three significant figures): So KBr is 119 g/mole
There are two steps to solving this problem:Step 1: Calculate the molar mass of KBr (molar mass is the amount, in grams, that one mole of a substance weighs).From the periodic table, we see that the molar mass of K is 39.10 g/mol (its average atomic mass). Similarly, the molar mass of Br is 79.90 g/mol. Adding these numbers together gives the molar mass of KBr, 119 g/mol.Step 2: Calculate the weight, in grams, using the molar mass and the number of moles.For every one mole of KBr, you have 119 grams of KBr:mass (m) = 119 g/mol * 3.30 molmass (m) = 392.7 gTherefore, 3.30 moles of KBr weighs approximately 392.7 grams.
119 grams of ice would produce 119 grams of liquid water when melted because the mass remains the same during a phase change.
The mass in grams of KBr contained in 2.50 moles of the compound is 2.5 times the sum of the gram atomic masses of hydrogen and bromine, or 2.50(1.008 + 79.904) or 202 grams, to the justified number of significant digits.
To calculate the grams of potassium bromite in 0.280 moles, you would multiply the number of moles by the molar mass of KBrO2, which is 119 g/mol. Therefore, 0.280 moles of KBrO2 would be 0.280 moles * 119 g/mol = 33.32 grams.
The mass of o,5 mole of uranium is 119,014455 g.
The mass number of potassium is about 39. The mass number of bromine is about 80. Total is 119. 119* 2.5 = 297.5. So 297.5 grams of potassium bromide has got 2.5 gram moles.
To find the grams of KBr in the solution, first calculate the moles of KBr present by using the molarity formula: moles = Molarity x Volume (L). Then, convert moles of KBr to grams using its molar mass. For KBr, the molar mass is approximately 119 g/mol. Finally, perform the calculation to find the grams present in the solution.
Easy...the periodic table gives the average mole weight of each element averaged out over all it's naturally occurring, if any, isotopes, i.e., K is 39.09 Br is 79.9 Add them up and it's (to three significant figures): So KBr is 119 g/mole
There are two steps to solving this problem:Step 1: Calculate the molar mass of KBr (molar mass is the amount, in grams, that one mole of a substance weighs).From the periodic table, we see that the molar mass of K is 39.10 g/mol (its average atomic mass). Similarly, the molar mass of Br is 79.90 g/mol. Adding these numbers together gives the molar mass of KBr, 119 g/mol.Step 2: Calculate the weight, in grams, using the molar mass and the number of moles.For every one mole of KBr, you have 119 grams of KBr:mass (m) = 119 g/mol * 3.30 molmass (m) = 392.7 gTherefore, 3.30 moles of KBr weighs approximately 392.7 grams.
119 grams of ice would produce 119 grams of liquid water when melted because the mass remains the same during a phase change.
16 grams of sugar.
119 grams = 4.2 ounces.
4.2oz = 119.07g
The mass in grams of KBr contained in 2.50 moles of the compound is 2.5 times the sum of the gram atomic masses of hydrogen and bromine, or 2.50(1.008 + 79.904) or 202 grams, to the justified number of significant digits.
It is approximately 1/2 a cup.