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Varvara68 [4.7K]
4 years ago
9

Consider two solutions, one formed by adding 10 g of glucose (c6h12o6) to 1 l of water and the other formed by adding 10 g of su

crose (c12h22o11) to 1 l of water. are the vapor pressures over the two solutions the same?
Chemistry
1 answer:
Jlenok [28]4 years ago
3 0

Relative lowering of vapour pressure is a colligative property

The colligative property depends upon the number of solute particles.

More the number of solute (non volatile) particles more the relative lowering of vapour pressure of solution.

The molar mass of sucrose is higher than the molar mass of glucose

thus for each gram of glucose and sucrose the moles of glucose will be more than the moles of sucrose.

hence for 10g of both glucose and sucrose

the moles of glucose will be more than moles of sucrose

Due to more number of molecules of non volatile solute in the solution of 10g of glucose the lowering of vapour pressure will be more in glucose solution.

hence the vapour pressure over glucose solution will be less than the vapour pressure over sucrose solution.

Answer: No,the vapor pressures over the two solutions will not be the same

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The human body needs at least 1.03 x 10-^2 mol O2 every minute. If all of this oxygen is used for the
Neko [114]
<h3>Answer:</h3>

0.3093 g of glucose are consumed each minute by the body.

<h3>Explanation:</h3>
  • During cellular respiration glucose is broken down in presence of oxygen to yield energy, water and carbon dioxide.
  • The equation for the reaction taking place during cellular respiration is;

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

We are required to calculate the amount of glucose in grams;

<h3>Step 1: Calculate the moles of glucose broken down</h3>

From the equation, the mole ratio of glucose to Oxygen is 1 : 6

Moles of Oxygen in a minute is 1.03 × 10^-2 moles

Therefore, moles of glucose will be;

= (1.03 × 10^-2)÷6

= 1.717 × 10^-3 moles

<h3>Step 2: Mass of glucose </h3>

Mass is given by multiplying the number of moles with molar mass

mass = moles × molar mass

Molar mass glucose is 180.156 g/mol

Therefore;

Mass = 1.717 × 10^-3 × 180.156 g/mol

         = 0.3093 g

Hence, 0.3093 g of glucose are consumed each minute by the body.

3 0
4 years ago
How do I go from molecules to moles? For example 1.50e23 molecules NH3 how many moles are in this problem?
maxonik [38]
To go from molecules to moles divide by Avogadro's number [<span>6.02x10^23]</span>
Example:
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4 0
3 years ago
A buffer solution contains 0.496 M KHCO3 and 0.340 M K2CO3. If 0.0585 moles of potassium hydroxide are added to 250. mL of this
aev [14]

Answer:

pH = 10.9

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:

n_{CO_3^{2-}}=0.34mol/L*0.250L+0.0585mol=0.1435mol\\\\n_{HCO_3^{-}}=0.34mol/L*0.250L-0.0585mol=0.0265mol

The resulting concentrations are:

[CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M \\

[HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M

Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:

pH=10.2+log(\frac{0.574M}{0.106M} )\\\\pH=10.9

Which makes sense since basic OH⁻ ions were added.

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6 0
4 years ago
Paragraph 2 discusses two different types of thermochemical reactions. In one type of
astraxan [27]

Answer: The reaction is endothermic.

Explanation:

Endothermic reaction : It is a type of chemical reaction where the energy is absorbed from the surrounding. In the endothermic reaction, the reactant have less energy than the energy of products

In endothermic reaction, enthalpy terms is located on the reactant side.

Example: A+B+energy\rightarrow C+D

Exothermic reaction : It is a type of chemical reaction where the energy is released into the surrounding. In the exothermic reaction, the energy of reactants are more than the energy of products

In exothermic reaction, enthalpy terms is located on the product side.

Example: A+B\rightarrow C+D+energy

8 0
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