volume of hydrogen gas generated when 3mL of 1M sodium borohydride reacts with concentrated hcl?
927283
NaBH4 + 2H2O → NaBO2 + 4H2\n\nthe moles of NaBH4 3*10^-3 * 1 = 3*10^-3 \nthe moles of H2 : 4*3*10^-3 = 12*10^-3\nthe volume of H2 : 12*10^-3 * 22.4 = 0.2688L\n\n2HCl + Zn => ZnCl2 + H2\n\nthe moles of Zinc : 1/65 moles\nthe moles of H2 : 1/65 moles\nthe volume of H2 : 1/65 * 22.4 = 0.3446 L
952447
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).
779566
17.13 mL = 0.01713 L\n\n0.01713 L x 0.3150 M = 0.005396 mol\n\nNaOH is a strong base, HCl is a strong acid.\nEach NaOH releases one hydroxide anion.\nEach HCl releases one proton.\n\nTherefore it takes 0.005396 mol of H+ to neutralize 0.005396 mol of -OH.\n\n0.005396 mol / 0.025 L = 0.2158 M\n\nThe concentration of the NaOH solution is 0.2158 M.
942319
This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)
598989
The amount of hydrogen by mass in water is 1/9.\nLiquid hydrogen has a density of roughly 1/14 that of water.\nThus the volume of liquid hydrogen would be 14/9 gallons, i.e. more by volume than you started with (Sounds wrong but it's true).\nIf you want me to convert it into terms of weight you'll have to use SI units. I always though a gallon was a warship they had back 400 years ago.
68089
With chemicals like NaBH4 you ought to say which solvent you mean. I will assume slightly alcaline water, adding OH(1-) to the equation:\n\nAg(1+) + 1 electron -> Ag (+-0)\n\nBH4(1-) -> "B(3+)" + "4 H(1+)" + 8 electrons.\n\nBoth the resulting hydrogen and the boron need hydroxide ions to remain in the solution, one each for every positive charge above. \n\nMultiply the silver equation by 8 to balance the electrons in the redox reaction.\n\nThis yields:\n\n8 AgNO3 + NaBH4 + 7 OH(1-) -> 8 Ag + 8 NO3(1-) + 4 H2O + H3BO3 + Na(1+)\n\nIf you need to write the result as neutral salts, add more sodium ions for the hydroxide.
1151475
number of moles of hydrogen is 0.2\n\namount of water is 2.4
910262
There is no sodium hydroxide. One mole of H2SO4 will react with 2 moles of NaOH in a neutralization reaction.
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
1116694
the reaction leads to production of hydrogen gas ...\n\n\n2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
367193
Draino is essential NaOH add that to aluminum and you will get NaOH + Al --> Al(OH)3 + Na2O + H2. So I guess hydrogen gas.
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
440307
Sodium Hydride (NaH) is an ionic compound that readily reacts with water or an alcohol to form hydrogen gas (H2).\n\nNaH --> Na+ (sodium ion) and H- (hydride) \n\nH2O donates an H+, with combines with the H- to form hydrogen gas, H2, a neutral molecule which will bubble out of the solution, leaving behing hydroxide ions from the water and sodium ions. In alcohol, the OH group donates the proton (H+), and is left with an O- where the alcohol group was located. The hydrogen gas will again bubble out of solution. \n\nHydrogen gas is highly flamable. In light of that fact, electolosis doesn't seem to be such a great technique to use with NaH.
259405
Total volume =( 1 + 2 ) dm3 = 3 dm3.\nUse Boyles  pV = constant on both gasses.\nThe partial pressure of helium is 2/3 kPa.\nThe partial pressure of neon is also 2/3 kPa.\nDalton's law:\nthe pressure of the mix is 2/3 kPa + 2/3 kPa = 1.3 kPa.
326922
Another easy reaction:\n\nMg(s) + HNO3 ===========> Mg(NO3)2 + H2 (g)\n\nMagnesium reacts rapidly with nitric acid to make magnesium nitrate and hydrogen gas
693257
From your reaction equation, 3 moles of hydrogen gas produces 2 moles of ammonia if the reaction was 100% efficient.  From the way you phrased your question, it's difficult to tell if you mean 450g of hydrogen (H), or 450 grams of hydrogen gas (H2).  I'll assume you mean hydrogen gas.\n\nFirst you have to calculate the number of moles hydrogen gas you started with:\n\nMoles H2 = grams H2 / Molecular weight H2\nMoles H2= 450 g H2 / 2.016 g per mole H2\nMoles H2 =223.21 moles\n\nIf the reaction was 100% efficient:\nMoles NH3 = 2/3 * moles H2\nMoles NH3 = 2/3 * 223.21\nMoles NH3 = 148.80 moles \n\nNow to calculate the grams NH3 expected from 100% efficiency:\ngrams NH3 = moles NH3 at 100% efficiency * mol wt NH3\ngrams NH3 = 148.80 moles * 17.024 grams/mole\ngrams NH3 = 2533.17 grams\n\nSo to calculate the percent yield for the reaction:\n\n% yield = (actual yield/ theoretical yield) *100\n% yield = (1575 grams/ 2533.17 grams) *100\n% yield = 62.17%\n\nNow if you meant H instead of H2 for the original grams hydrogen in your question, you'll have to recalculate as appropriate.