Chemistry 12 Thought lab Part 1 (Procedure) 1) 1s22s22p23s23p24s23d24p25s24d25p26s24f25d26p2 2) When ml can only equal 1‚ each energy level can have only 1 orientation so according to the exclusion principle only 2 electrons with opposite spins can be in each orbital. So in order to get to element 30 you would need to go all the way to the 6p orbitals. In other words‚ every two electrons would necessitate going to a new orbital. 1 | 1s1 | 11 | 1s2 2s2 2p2 3s2 3p2 4s1 | 21 | 1s2 2s2 2p2 3s2 3p2
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Mole catcher by: Sam winters A mole catcher is a spy in the war. The second Continental congress created a Secret committee by a resolution on september 18‚1775. The committee was not a true intelligence agency. Since the committee of secret correspondenceoften worked was mainly concerned with obtaining military supplies in secret and distrubting them‚ and selling gun powder previously negotiated by certain members of the congress without the formal sanction of that body.. the committee kept its
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Limiting Reagents and Percentage Yield Worksheet 1. Consider the reaction I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g) a) 80.0 grams of iodine(V) oxide‚ I2O5‚ reacts with 28.0 grams of carbon monoxide‚ CO. Determine the mass of iodine I2‚ which could be produced? 80 g I2O5 1 mol I2O5 1 mol I2 1 333.8 g I2O5 1 mol I2O5 28 g CO 1 mol CO 1 mol I2 253.8 g I2 1 28 g CO 5 mol CO 1 mol I2 b) If‚ in the above situation‚ only 0.160 moles‚ of iodine‚ I2 was produced
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Homework has traditionally been considered a reinforcement of what students learn throughout the school day and is often used as a tool to make sure students have fully understood the curriculum at hand. Excessive amounts of homework constantly put pressure on students and cause them to spend outrageous amounts of time taking away from their social lives‚ families‚ friends‚ and extra-curricular activities as well as sports. Excessive amounts of homework can also cause students to stay up till midnight
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[pic] Basics of Economics Submitted to: Ms. Deepa Bhaskaran Submitted By: Sahil Bansal Venkat Batra Saurabh Satija Ravikant Gupta NANO CASE Q-1 Identify target market for nano ? Ans- The target market for nano would be The group of people that belongs th the lower-middle class
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acid‚ CH3CHOHCOOH‚ in an excess of O2? 3. 31.5 g of Na2S2O3 was made to react with 18.6 g of Cl2 according to the balanced chemical reaction below: Na2S2O3(aq) + 4Cl2(g) + 5H2O(aq) → 2NaHSO4(aq) + 8HCl(aq) a. Which is the limiting reactant? b. What mass (in g) of the excess reactant remains after the reaction? c. What is the theoretical yield (in g) of NaHSO4? IV. FINAL BOSS! A 25.00 g sample of deep-sea sediment was collected and it was found that 12.56% of it was equivalent to the weight (in g)
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formula CuSO4.H2O‚ we need to find the ratio of mole between anhydrous copper II sulfate and the water vapour by using the formula: In this case‚ the molar mass of: Copper II sulfate (CuSO4) is: 63.55 + 32.06 + 15.99 x 4 = 159.57 g/mol Water (H2O) is: 1.01 x 2 + 15.99 = 18.01 g/mol (e.g. 0.37/18.01= 0.02054 mol.) Then‚ is obtained by: Calculation 1st attempt 2nd attempt 3rd attempt Mole of water (mol) = 0.06163 Mole of anhydrous copper sulfate (CuSO4)
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SUBMISSION INSTRUCTIONS: WHAT: Printed questionnaire with the answers written in a big yellow booklet. One page per number. Paginate each page accordingly. Use different inks to distinguish the bonds of electrons for numbers involving molecular structure. WHO: All students with Student Number ending in ODD number will answer Set A while students with Student Number ending in EVEN number will answer Set B. Indicate your Student Number and the Set you are answering in front of the yellow
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Question 1 1 out of 1 points The volume of a regular cylinder is V = πr2h. Using the value 3.1416 for the constant π‚ the volume (cm3) of a cylinder of radius 2.34 cm and height 19.91 expressed to the correct number of significant figures is _________. Selected Answer: d. 342 Correct Answer: d. 342 Response Feedback: Correct Question 2 1 out of 1 points There are ____________ significant figures in the answer to the following computation: (29.2 - 20.0 ) (1.79 x 105)
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6e- 7e- ^ the 13 on C is mass number. = # protons + # neutrons. 3. N2 + 2 O2 + Cl2 -> 2 NO2 Start with 6 mol N2‚ 4 mol O2‚ 4 mol Cl2‚ Find limiting reagent and amounts remaining of excess reactants N2 + 2 O2 + Cl 2 6 mol 4 mol 4 mol start Max yield x (2mol No‚Cl/ 1 mol N2) Of No2Cl = 12 mol 4 mol 8 mol O2 is the limiting reagent. 4 mol of NO2Cl (4 mol NO2Cl) (1 mol N2 / 2mol No2Cl) = 2 mol N2 consumed 6 mol N2 – 2 mol N2 = 4 mol N2 left ^at start ^ used up (4 mol NO2Cl)
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