Are there a mole of golf in a 200 g gold necklace

What happens in the body when we break down carbohydrates? Explain the whole course of events.

JulsSmile [24]

Answer:

Energetic molecules are formed, or rather the energetic currency to be able to live, which we call ATP.

Explanation:

ATP is a molecule that arises from the degradation of carbohydrates, these are destroyed for the first time in the mouth by the enzyme salivary amylase, then by the stomach, and finally they are absorbed in the intestine in the form of glucose, this glucose enters the bloodstream, entering the cells through insulin.

Once glucose is entered into the cell, it completes an energetic cycle, called the krebs cycle, which provides 36 to 38 ATP.

Atp, is adenosine tri phosphate, phosphate bonds protect a lot of energy, this is how the body generates the following reaction when it needs energy, releasing energy to the environment to be able to produce from locomotion to respiration:

ATP -----> ADP + Pi

8 0

3 years ago

If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so

Leviafan [203]

Answer:

Final pH: 9.49.

Round to two decimal places as in the question: 9.5.

Explanation:

The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.

What's the pKb of base B?

Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.

$\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}$ .

$\text{pOH} = \text{pK}_w - \text{pH}$ .

$\text{pK}_w = 14$ .

$\text{pOH} = 14 - 9.5 = 4.5$

$\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613$ .

What's the new salt-to-base ratio?

The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.

Initial:

$n(\text{B}) = c\cdot V = 1.05 \times 0.5 = 0.525\;\text{mol}$ ;
$n(\text{BH}^{+}) = c\cdot V = 0.750 \times 0.5 = 0.375\;\text{mol}$ .

After adding the HCl:

$n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol}$ ;
$n(\text{BH}^{+}) = 0.375+ 0.005 = 0.380\;\text{mol}$ .

Assume that the volume is still 0.5 L:

$\displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}$ .
$\displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}$ .

What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

$\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991$

$\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49$ .

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

3 0

3 years ago

9. How much power is flowing into your house if the voltage is 120 V and the current is 50 A? Use the formula P = VI.

lyudmila [28]

4800 mannnnnnnnnnnnnn

4 0

3 years ago

Balance the below equation:

svet-max [94.6K]

Answer:

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

This is an oxidation-reduction (redox) reaction:

4 N-III - 20 e- → 4 NII

(oxidation)

10 O0 + 20 e- → 10 O-II

(reduction)

NH3 is a reducing agent, O2 is an oxidizing agent.

Explanation:

good luck

6 0

3 years ago

How does density affect water circulation?

Kipish [7]

Answer:

Im guessing it's C....................................................

4 0

3 years ago

Read 2 more answers