Why is the formation of fructose-1,6-bisphosphate a step in which control is likely to be exercised in the glycolytic pathway

What is the pH of a 1.9 M solution of HClO4?

Sliva [168]

The compound HClO4, when placed in water, will dissociate into the ions, H+ and ClO4-. Therefore, the 2.0 M solution will also form 2.0 M H+. The pH is calculated through the equation,
pH = -log[H+]
Substituting,
pH = -log[2] = -0.3
Thus, the pH of the solution is -0.3.

5 0

2 years ago

An average microwave uses a frequency of 2.54 GHz. How much energy is emitted?

romanna [79]

Answer:

0.120 m

Explanation:

What you need to know here is that frequency and wavelength have an inverse relationship as described by the equation

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4 0

2 years ago

Determine the concentration of a solution made by dissolving 1.40grams NaCL in enough water to make 30.0mL of solution.

KiRa [710]

Answer: 0.8M

Explanation:

Given that,

Amount of moles of NaCl (n) = ?

Mass of NaCl in grams = 1.40 g

For molar mass of NaCl, use the molar masses:

Sodium, Na = 23g;

Chlorine, Cl = 35.5g

NaCl = (23g + 35.5g)

= 58.5g/mol

Since, amount of moles = mass in grams / molar mass

n = 1.40g / 58.5g/mol

n = 0.024 mole

Now, given that:

Amount of moles of NaCl (n) = 0.024

Volume of NaCl solution (v) = 30.0mL

[Convert 30.0mL to liters

If 1000 mL = 1L

30.0mL = 30.0/1000 = 0.03L]

Concentration of NaCl solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

c = 0.024 mole / 0.03 L

c = 0.8 M (0.8M means concentration is in moles per litres)

Thus, the concentration of the solution is 0.8M

3 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago