PERCENT WATER IN A HYDRATE DATA/CALCULATION SHEET NAME 1. crucible, cover, BaCl2 2H2O \8,ao6! crucible and cover 14.0188 g BaCl, 2H2O 4.1873 first second third weighing weighing weighing crucible, cover, anhydrous BaCl, 7.5824 17.5815 H2O lost (use your last weighing) 7.58158 % water, experimental SHOW CALCULATION

5.71 g of nitrogen reacts with a certain amount of hydrogen according to the following reaction: 3 H2 + N2 → 2 NH3 How many grams of hydrogen would be required to react with 5.71 g of nitrogen ? Select ALL the conversion factors needed for this calculation using dimensional analysis method.   1) 1 mol N2/3mol H2 2) 28.02 g N2/ 1mol N2 3)1 mol H2/2.016 g H2 4)2.016 g H2/1mol H2 5)1 mol N2/28.02 g N2 6) 3 mol H21 mol N2

How would i solve

mistry app.101edu.co YouTube Maps NG AMOUNT * @ 7 2 X W S F2 X Jill # Welcome to MyTCC Welcome to MyTCC 3 F3 E D C 20: $ 4 R F 30 % 35.6 V Aluminium has a density of 2.70 g/cm² How many moles of aluminum are in a 13.2 cm block of the metal substance? 4 ADD FACTOR X x() F5 5 molecules Al T Content G 1.32 F6 A 6 B 2.70 0.489 g Al/mol F7 Y & H Question 1 of 50 0.100 g/cm² W 7 26.98 N ASUS X F8 U J g Al Lo b Home | bartleby ANSWER 8 962 13.2 F9 | M mol Al 1 1 K F10 9 < I RESET 5 6.022 x 10 O cm 1 F11 O L X + 10 P : F12 - I Prt Sc 1 ? + [ Q 12 ☆ = 12 Insert 1 Delete Backspace J Submit Home Enter PgUp Shift Pe

Esc STARTING AMOUNT 71°F Sunny Z²² F1 DX X F2 A- F3 F4 According to the balanced reaction below, calculate the quantity of moles of NH, gas that form when 4.20 mol of N₂H4 liquid completely reacts: 4.20 5.60 ▬▬ *- ADD FACTOR x( ) F5 17.04 1.40 g N₂ *+ O Search the web F6 3.15 3.37 x 1024 mol N₂H₁ 3 N₂H.(1)→ 4 NH.(g) + N2(g) ► / || F7 Question 9 of 11 2.53 x 1024 32.06 4 g NH, F8 12.6 g N₂H4 28.02 ANSWER 1.05 4 pity mol NH, F9 1 6.022 x 1023 8.43 x 102 3 * RESET mol N₂ 2 16.8 F10 Insert F11 F12 Delete

Can you please solve using box method?

Empirical Formula for Unknown Sample Data1. Mass of crucible 28.072(grams)2. Mass of crucible + Unknown 30.528(grams)3. Mass of crucible + Na + S 29.782(grams)4. Mass of crucible + Na 28.786(grams)5. Mass of Na 0.714 6. Mass of S 0.996 7. Mass of O 0.7468. Mass of Unknown Sample 2.51 using the information above find the moles for the following elements: Moles of Na Moles of S Moles of O

Conc. of NaOH: TRIAL 1 TRIAL 2 TRIAL 3 Mass of flask 81.061 69.692 96.039 Mass of flask and vinegar 83.010 71.578 97.940 Mass of vinegar 1.949 1.886 1.901 Volume of vinegar 2.00 2.00 2.00 Initial buret reading Final buret reading 17.0 17.5 16.0

3 Answer in General Chemistry fo X A app.101edu.co ps M Gmail YouTube Maps Tradůzir GE Notícias Question 12 of 15 What mass in grams of NaN3 is required to produce 50.2 L of N2 gas (density = 1.25 g/L ) according to the balanced chemical reaction: 2 NaN3(s) → 2 Na(s) + 3 N2(g) STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) %3D 2 3 97.1 77.7 22.99 1 1.25 6.022 x 1023 65.02 50.2 28.02 146 218 g N2/L mol N2 mol Na mol NaN3 g N2 g NaN3 g Na L N2 山T 286 APR 1 PAGES W 17 MacBook Air ....

Data Trial 1 Trial 2 Mass of empty crucible with lid 26.698g 26.687g Mass of Mg metal, crucible, and lid 27.060g 27.046g Mass of MgO, crucible, and lid 27.291g 27.273g Determine the percent yield of MgO for your experiment for each trial.   Trial 1:98.8% Trial 2:98.3% (Just write out how you got both percent yields and follow signifigant figure rules)