Here they are : 1ST VERSE G G G G E G E G E G E G G G G G G G G E G E G E G E G G G A G G HIGHD B A B A B A B A G G G G G G G G E G E G E G E G G G G G G G A G G G G G G G A G B A G CHORUS B B B B A G A B D E G A G E G A G E B B B B A G A B D E G A G E G A G E G G G G G G G G G G A B G G G G A F G A B D E Please add more if you know the rest.
G A B D E D A G A B D E G G D E G A B D E D A G A B D E G G D E G Fs G G G A G Fs G G G G A G Fs G G G D C G Fs G G G A G Fs G G G G A G Fs G G G D C G A G A B A G G A G A B G A G E G A G E G A G A B
G A B D E D A G A B D E G G D E G A B D E D A G A B D E G G D E G Fs G G G A G Fs G G G G A G Fs G G G D C G Fs G G G A G Fs G G G G A G Fs G G G D C G A G A B A G G A G A B G A G E G A G E G A G A B A G G A G A B G A G E G A G E E D.(X2)
Baby by: Justin BieberF G F F A G F E D E D F A G F E D E D F A G F G G F EF C2 A G A F C2 A G F C2 A G A F C2 A G F C2 C2 A G F C2 C2 A GF F A G A G A G A G A G F C2 A G A C2 A G F C2 G A F F G F F F A A G F GF F G G G G G G A G F F G FChorus: A G A G A G C2 G A G A G A G D2 G A G A G A G C2 A G A A A G FA G A G A G C2 G A G A G A G D2 G A G A G A G C2 A G A A A G F-Rossele-Send more requests @ycel_gandah@yahoo.comTy!
easy-g a g g g g a g g g a g a a (2 times)-b b g g a a b g bb g g a a g -g a g g g g a g g g a g a a (2 times) hard-g a b b c b a g b b a g a a g a b b c b a g b b a g g a g d d b b a b c a b c d d a b c a b
Divide 6.10 (g NaHCO3) by 84.007 (g.mol−1 NaHCO3) to get 0.0726 mol NaHCO3
The mass of oxygen is o,303 g.
NaHCO3(aq) + CaHPO4(aq) → NaCaPO4(aq) + CO2(g) + H2O(l)
For this you need the atomic (molecular) mass of NaHCO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaHCO3=84.0 grams110 grams NaHCO3 / (84.0 grams) = 1.31 moles NaHCO3
12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HClneeds62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3
NaHCO3 ----> H2O Mass 2.10g 0.045g RAM 84 g/moles 18 g/moles number of moles 0.025moles 0.025moles
Sodium Bicarbonate come from absorption of CO2 into NaOH Total reaction is 2NaOH + CO2 -> NaHCO3 + H2O Each mole of NaHCO3 come from absorption of 1 mole CO2 Molecular weight of NaHCO3 is 84 g/mol and CO2 is 44 g/mol It is that 84 g of NaHCO3 had 44 g of CO2 By weight ratio it is 52% CO2 in Sodium Bicarbonate.
3.00 grams NaHCO3 (1 mole/84.008 grams) = 0.0357 moles of sodium bicarbonate
1 mole of NaHCO3 Na = 1 * 22.99 g = 22.99 g H = 1 * 1.01 g = 1.01 g C = 1 * 12.01 g = 12.01 g O = 3 * 16.00 g = 48.00 g Total = 84.01 g There are 84.01 grams in one mole of NaHCO3.
The answer to the conversion is that 35.0 grams of hydrochloride (HCL) equals 0.76 moles. The conversion rate is 35.0 grams divided by 46 gram per mole. A mole is the molecular weight of a substance.
500 mL * 100(mMol/mL) = 50 mMol NaHCO3 , hence50 mMol NaHCO3 = 50(mMol) * 84(mg/mMol) = 4200 mg = 4.2 g NaHCO3 in 500 mL
NaHCO3 = 84.007 g/mol Figure out the mole fraction = 2g / 84.007 g/mol The unit of the answer is mol, which in this context is the same as mole. Since you only have one carbon in NaHCO3 you cannot have more moles of CO2 than moles of NaHCO3.