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3 years ago

Please help this is due tonight

Chemistry

1 answer:

Elenna [48]3 years ago

5 0

4. the nervous system uses electrical impulses to send messages through neurons while endocrine glands use hormones to send messages

5. Hormones are released from the endocrine glands

6. a. neck
b. Chest

7. help regulate important functions, such as growth, blood pressure and reproduction; it is small and small and bean shaped.

8. produces hormones that regulate the body's metabolic rate controlling heart, muscle and digestive function, brain development.

9. regulate your metabolism, immune system, blood pressure, respond to stress etc.

10. “enzymes” - break down sugars, fats, and starches

11. cells are unable to use glucose (can’t break down sugar to use as energy)

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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

Can someone help me with thins, im to lazy to read rn i did freaking 47 over due homeworkkkk

AysviL [449]

Smh das tough and all and same i got hella overdue

3 0

3 years ago

Water (2350 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the ini

Eva8 [605]

Answer:

40.7062 °C

Explanation:

Let the initial temperature = x °C

Boiling temperature of water = 100 °C

Using,

Q = m C ×ΔT

Where,

Q is the heat absorbed in the temperature change from x °C to 100 °C.

C gas is the specific heat of the water = 4.184 J/g °C

m is the mass of water

ΔT = (100 - x) °C

Given,

Mass = 2350 g

Q = 5.83 × 10⁵ J

Applying the values as:

Q = m C ×ΔT

5.83 × 10⁵ = 2350 × 4.184 × (100 - x)

<u>x, Initial temperature = 40.7062 °C </u>

3 0

3 years ago

which of the following will stay consant, no matter if the substance is in the solid, liquid, or gas state

777dan777 [17]

Answer:

chemical composition

Explanation:

changes into any state of matter are physical and thus the chemical composition of the matter is unaltered

8 0

2 years ago

11. Sound's loudness (or volume) is measured in decisels,<br> A True<br> B. False

bagirrra123 [75]

Answer:

sound loudness is measured in decibels, A. True

7 0

3 years ago