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son4ous [18]
3 years ago
11

solid sodium reacts violently with water producing heat, hydrogen gas and sodium hydroxide. How many molecules of hydrogen gas a

re produced when 65.4g of sodium are added to water? 2Na(s)+ 2H2O(l) -->. 2NaOH(aq) + H2(g)
Chemistry
1 answer:
Blababa [14]3 years ago
5 0
Molar mass of sodium=23g
2 mole sodium gives 2 g H2
i.e 46g sodium gives 2g H2
so 65.4g will give=2.84g H2
now no. of molecules = 2.8/2*avogadros number
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Color.
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Volume.
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Hope this helps. :)
4 0
3 years ago
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe
polet [3.4K]
First, we need the no.of moles of O2 = mass/molar mass of O2
                                                             = 55 g / 32 g/mol
                                                             = 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2 
So we can get the no.of moles of H2O = 2 * moles of O2
                                                                  = 2 * 1.72 mol
                                                                  = 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT

when P = 12.4 atm  & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
 ∴ V ≈ 8.2 L 
4 0
3 years ago
An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm
frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0
3 years ago
A 1 mL pipette delivers a measured volume of .98 mL (calculate the percentage error)
Natalija [7]

Answer:

2%

Explanation:

.98 is 98% of one and therefore they are missing 2%

4 0
3 years ago
Read 2 more answers
A sample of 19 mg of 9-fluorenone is dissolved in 5 mL of diethyl ether. The resulting organic
maks197457 [2]

Answer:

15.70mg would remain

Explanation:

Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:

Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O

After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:

(19mg - X)

<em>Where X is the mass that now is in the aqueous phase</em>

Replacing in Kp formula:

9.5 = (19mg - X) / 5mL / (X /10mL)

0.95X = 19mg - X / 5mL

4.75X = 19 - X

5.75X = 19

X = 19 / 5.75

X = 3.30mg

That means 9-fluorenone that remain in the ether layer is:

19mg - 3.30mg =

<h3>15.70mg would remain</h3>
8 0
3 years ago
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