t Bi 2 O 3 l TiO 2 n BaCO 3 p Na 2 CO 3

Sep 01, 2014

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Stages in the formation of NBT-7%BT. NBT-7BT 900°C c alcination R p 0.0696. t Bi 2 O 3 l TiO 2 n BaCO 3 p Na 2 CO 3. 100. 110. 111.

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t Bi 2 O 3 l TiO 2 n BaCO 3 p Na 2 CO 3

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Stages in the formation of NBT-7%BT NBT-7BT 900°C calcination Rp 0.0696 t Bi2O3 l TiO2 n BaCO3 p Na2CO3 100 110 111

1) 2) CaCl 2 • 2H 2 O(aq) + Na 2 CO 3 (aq)  2H 2 O + CaCO 3 (s) + 2NaCl(aq) 3)

0.00943 mol Na 2 CO 3. = . 1 mol CaCl 2 • 2H 2 O. 0.0136 mol CaCl 2 • 2H 2 O. x . = . 147.02 g CaCl 2 • 2H 2 O. 1 mol Na 2 CO 3. x . 105.99 g Na 2 CO 3. # mol CaCl 2 • 2H 2 O =. 2.00 g CaCl 2 • 2H 2 O. 1) 2) CaCl 2 • 2H 2 O(aq) + Na 2 CO 3 (aq)

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K 3 [Fe(CN) 6 ] K 2 [PtCl 4 ] Na 2 [Fe(CO) 4 ] [Co(H 2 O) 2 (NH 3 ) 4 ]Cl 3 [Ni(H 2 O)(NH 3 ) 4 ]SO 4 Na 2 [OsCl 5 N] [C

NOW some for you to try!!!. K 3 [Fe(CN) 6 ] K 2 [PtCl 4 ] Na 2 [Fe(CO) 4 ] [Co(H 2 O) 2 (NH 3 ) 4 ]Cl 3 [Ni(H 2 O)(NH 3 ) 4 ]SO 4 Na 2 [OsCl 5 N] [CoCl(NO 2 )(NH 3 ) 4 ]Cl [CoCl(NH 2 )(en) 2 ]NO 3 [FeH(CO) 3 (NO)] [PtCl(NH 2 CH 3 )(NH 3 )]Cl. Potassium hexacyanoferrate(III)

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L:, 2:, 3:ALKH

263 views • 10 slides

2 Co.3:18

2 Co.3:18. We all : every believer Unveiled face : looks at Lord w/o veil, unlike Moses (covered face before Israel) Behold ... and reflect : Jn.1:14; 2 Pt.1:16-17; Ac.4:13; 1 Pt.2:21 Transformed : Mt.17:2. Moses reflected God’s glory; our face reflects glory

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H 2 + N 2 NH 3

H 2 + N 2 NH 3. 3H 2 + N 2 2NH 3. 3H 2 + N 2 2NH 3. +. 3H 2 + N 2 2NH 3. 3H 2 + N 2. 3H 2 + N 2. N 2 + 3H 2 2NH 3. 9H 2 + 3N 2 6NH 3. Stoichiometry:. How and why can a chemical equation give us information for calculating quantities

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H 2 CO 3

Early Proximal tubule. Lumen. Blood vessel. Na +. Na +. Na +. ATP. Na +. Na +. Na +. Na +. (65 -80). Na +. Na +. Na +. Na +. K +. Glucose . Na +. Na +. Na +. Na +. Na +. Na +. Na +. Na +. H +. HCO 3 -. H +. + . + . HCO 3 -. H +. HCO 3 -.

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[SO 2 + H 2 O   H 2 SO 3 ] SO 3 + H 2 O  H 2 SO 4 CO 2 +H 2 O  H 2 CO 3

[SO 2 + H 2 O   H 2 SO 3 ] SO 3 + H 2 O  H 2 SO 4 CO 2 +H 2 O  H 2 CO 3 2 NO 2 +H 2 O  HNO 2 + HNO 3. SO 2 + 2 HO-Radikale  H 2 SO 4 CO + 2 HO-Radikale  H 2 CO 3 NO + HO-Radikal  HNO 2 NO 2 + HO-Radikal  HNO 3.

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CO 2 i vann CO 2 + H 2 0  H 2 CO 3 H 2 CO 3 H + + HCO 3 - HCO 3 - H + + CO 3 2-

CO 2 i vann. CO 2 i vann CO 2 + H 2 0  H 2 CO 3 H 2 CO 3 H + + HCO 3 - HCO 3 - H + + CO 3 2- NaHCO 3 Na + + HCO 3 -. CO 2 i vann. CO 2 + H 2 0  H 2 CO 3 H 2 CO 3 H + + HCO 3 - HCO 3 - H + + CO 3 2- NaHCO 3 Na + + HCO 3 -.

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C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O

C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O. 220 g of C 3 H 8 react in the above reaction. What mass of H 2 O is produced?. Pb (NO 3 ) 2 + 2 NaI  PbI 2 + 2 NaNO 3. 50 mL of a 0.5 mol /L solution of NaI reacts to produce how many grams of PbI 2 ?. C =. n V.

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Li 2 CO 3 + O 2 + Heat = C0 2 + Li 2 O

Li 2 CO 3 + O 2 + Heat = C0 2 + Li 2 O. HCl + Cu = No Reaction. C 6 H 12 O 6 + H 2 SO 4 = H 2 O + C + ?. Waiting for the “expert.”. C 6 H 12 O 6 + KNO 3 + O 2 + Heat = CO 2 + H 2 O + C + K 2 O. H 2 SO 4 + Mg = H 2 + MgSO 4. PbNO 3 + KI = KNO 3 + PbI.

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Mg + N 2  Mg 3 N 2

Practice. Magnesium metal is burned in nitrogen. Mg + N 2  Mg 3 N 2. Lead foil is immersed in silver nitrate. Pb + Ag +  Pb 2+ + Ag. A solution of ammonium sulfate is added to a saturated solution of barium hydroxide. NH 4 + + SO 2+ + Ba 2+ + OH -  BaSO 4 + NH 3 + H 2 O.

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Warm Up Simplify. 1. 2(2) 2. (–2)(–2) 3. (–2)(–2)(–2) 4. 3(3)(3)

4 9. Warm Up Simplify. 1. 2(2) 2. (–2)(–2) 3. (–2)(–2)(–2) 4. 3(3)(3). 4. 4. –8. 27. 5. Objective. Evaluate expressions containing exponents. Vocabulary. power base exponent. 3. 2.

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B 2 O 3 + 3 H 2 O ---> 2 H 3 BO 3

B 2 O 3 + 3 H 2 O ---> 2 H 3 BO 3. If diboron trioxide is reacted with water, the product is boric acid. What mass of boric acid is obtained from the full reaction of 6.992 g of diboron trioxide?. 17 H 2 + 8 CO ---> C 8 H 18 + 8 H 2 O.

379 views • 26 slides

T ransition Metal Doped Sb 2 Te 3 and Bi 2 Te 3

Sb 2-x Cr x Te 3 50 nm. ferromagnetic, H C2. Easy axis. Sb 2 Te 3 0-10 nm. nonmagnetic. Sb 2-y Cr y Te 3 50 nm. ferromagnetic, H C1. Sapphire substrate. T ransition Metal Doped Sb 2 Te 3 and Bi 2 Te 3

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K 3 [Fe(CN) 6 ] K 2 [PtCl 4 ] Na 2 [Fe(CO) 4 ] [Co(H 2 O) 2 (NH 3 ) 4 ]Cl 3

433 views • 40 slides

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t Bi 2 O 3 l TiO 2 n BaCO 3 p Na 2 CO 3

t Bi 2 O 3 l TiO 2 n BaCO 3 p Na 2 CO 3

Presentation Transcript

1) 2) CaCl 2 • 2H 2 O(aq) + Na 2 CO 3 (aq)  2H 2 O + CaCO 3 (s) + 2NaCl(aq) 3)

K 3 [Fe(CN) 6 ] K 2 [PtCl 4 ] Na 2 [Fe(CO) 4 ] [Co(H 2 O) 2 (NH 3 ) 4 ]Cl 3 [Ni(H 2 O)(NH 3 ) 4 ]SO 4 Na 2 [OsCl 5 N] [C

L:, 2:, 3:ALKH

2 Co.3:18

H 2 + N 2 NH 3

H 2 CO 3

[SO 2 + H 2 O   H 2 SO 3 ] SO 3 + H 2 O  H 2 SO 4 CO 2 +H 2 O  H 2 CO 3

CO 2 i vann CO 2 + H 2 0  H 2 CO 3 H 2 CO 3 H + + HCO 3 - HCO 3 - H + + CO 3 2-

C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O

Li 2 CO 3 + O 2 + Heat = C0 2 + Li 2 O

Mg + N 2  Mg 3 N 2

Warm Up Simplify. 1. 2(2) 2. (–2)(–2) 3. (–2)(–2)(–2) 4. 3(3)(3)

B 2 O 3 + 3 H 2 O ---&gt; 2 H 3 BO 3

T ransition Metal Doped Sb 2 Te 3 and Bi 2 Te 3

K 3 [Fe(CN) 6 ] K 2 [PtCl 4 ] Na 2 [Fe(CO) 4 ] [Co(H 2 O) 2 (NH 3 ) 4 ]Cl 3

B 2 O 3 + 3 H 2 O ---> 2 H 3 BO 3