3^6 = 729
27 3 to the power of 3 3 x 3 x 3 (3 x 3) x 3 9 x 3 27
x/3
2 x 3 x 3 x 3 = 54 2 x 3 x 5 x 7 = 210 2 x 3 x 3 x 3 x 5 x 7 = 1890, the LCM
2 x 2 x 2 x 3 x 3 x 3 = 216 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288 2 x 2 x 2 x 3 x 3 x 5 = 360
20 x 500 mg = 10g or 10g = 500mg x 20
10g/410g X 100 = 2.4 2.5 %
By stoichiometry, every 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, 10g of N2 corresponds to 0.357 moles. Since there is excess H2, all 10g of N2 would react to produce NH3, which would be 0.476 moles (0.357 moles of N2 * 2 moles of NH3 / 1 mole of N2). The mass of NH3 produced would be 9.07g (0.476 moles * 17.03g/mol).
If you are on 10g or later and the recycle bin feature is turned on, you can use the FLASHBACK TABLE x TO BEFORE DROP command, where x is the name of the table. If it is not there, or the recycle bin feature is not available, you will have to restore the table from a backup.
5(X - 5) + 2g(X - 5)-------------------------------all the factoring I can see
To determine the number of atoms in 10g of Fe, you first need to calculate the number of moles of Fe using its molar mass (55.85 g/mol). Then, you can use Avogadro's number (6.022 x 10^23 atoms/mol) to find the number of atoms in those moles of Fe. Finally, multiply the number of moles by Avogadro's number to get the total number of atoms in 10g of Fe.
1 kg = 1000g (kilo=1000)0.01 kg = 0.01 x 1000g = 10g
10G networks transfer at a maximum of 10 gigabits per second.
To calculate the number of moles in 10g of iodine, you need to first determine the molar mass of iodine (I), which is 126.9 g/mol. Then, you can use the formula: moles = mass / molar mass. So, moles = 10g / 126.9 g/mol ≈ 0.079 moles of iodine.
After 2 half-lives (two half-lives of tritium is 12.32 x 2 = 24.64 years), the initial 10g sample of tritium would have decayed by half to 5g.
No, they do NOT contain the same number of particles (either molecules N2 or atoms C): there molecular or atomic mass is different.Look at this:10g N2 / 28 (g/mole N2) x 6.022x1023 (molecules N2/mole N2) = 2.15 x1023 molecules N210g C / 12 (g/mole C) x 6.022x1023 (atoms C/mole C) = 5.02 x1023 atoms C
10 grams nitrogen (1 mole N/14.01 grams)(6.022 X 1023/1 mole N) = 4.3 X 1023 atoms of nitrogen ======================