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3 years ago

What is the percent Hydrogen in glucose

Chemistry

1 answer:

Ivan3 years ago

7 0

Glucose- C6H12O6
12+6+6=24
H=12
12/24=0.5
0.5x100
=50%

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Can Magnesium hydroxide be used as the base for this titration? Why or why not? 5. Describe the procedure that you will follow t

kvasek [131]

Explanation:

Yes

2KHP + Mg(OH)2 = 2KOH + Mg(HP)2

KHP is an acidic salt which reacts with Mg(OH)2 which is a base via double displacement/decomposition reaction.

• One approach is to prepare the solution volumetrically using KHP crystals.

• Preparing 1 liter of 0.1 M KHP you would add 0.1 moles of KHP

Molar mass of KHP(C8H5KO4)

= (12*8) + (5*1) + 39 + (16*4)

= 204 g/mol

Number of moles = mass/molar mass

= 20.4/204

= 0.1 mol.

• This is added to a 1 liter volumetric flask, add deionized water until near the fill line, stopper and mix

Number of moles = molar concentration * volume

= 0.025 * 0.100

= 0.0025 mol

Equation for the reaction:

Ba(OH)2 + 2KHP --> Ba(KP)2 + 2H2O

1 mole of Ba(OH)2 reacted with 2 moles of KHP. By stoichiometry,

Number of moles of Ba(OH)2 = 0.0025/2

= 0.00125 mol.

Molarity = number of moles/volume

= 0.00125/0.01

= 0.125 M of Ba(OH)2.

4 0

3 years ago

How many sodium ions are in 3.0 moles of nacl

Elena L [17]

1 molecule of NaCl contains 1 sodium ion (Na+), that's why if we have 3.0 moles of.
NaCl, we have 3.0 moles of Na+.
N(ions) = n(mol) · NA.
N(ions) = 3.0 moles · 6.02·1023 = 18.06 ·1023 ions.

4 0

3 years ago

Read 2 more answers

Lenses are made to have different focal points. What two properties of a lens

Novay_Z [31]

The correct answer is letter C

3 0

3 years ago

the radius of magnesium atom is 0.160nm. the radius of a magnesium ion is 7.2x 10^-11m. explain the difference in size between t

Allushta [10]

Answer:

during reaction magnesium lises ions.

Explanation:

magnesium reacts by losing two ions which makes it smaller in size.

8 0

3 years ago

A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g

Lynna [10]

Answer:

$\%NaHCO_3=61.2\%$

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

$NaHCO_3+HA\rightarrow NaA+H_2O+CO_2$

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

$m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g$

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

$\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%$

Best regards.

5 0

3 years ago