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3 years ago

What is the [OH-] in a solution if the [H*] = 1.2 x 10-3 M?

Chemistry

1 answer:

spayn [35]3 years ago

7 0

We know that [OH⁻] * [H⁺] = 10⁻¹⁴

plugging the value of [H⁺]

[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ * (10³/1.2)

[OH⁻] = 833.3 * 10⁻¹⁴

[OH⁻] = 8.33 * 10⁻¹²

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For many purposes we can treat butane (C4H10) as an ideal gas at temperatures above

masha68 [24]

Answer:

A. Yes

B. –176 °C

Explanation:

A. Yes

B. Determination of the new temperature of the gas.

Let the initial pressure be P

From the question given above, the following data were obtained.

Initial pressure (P1) = P

Initial temperature (T1) = 19 °C

Final pressure (P2) = ⅓ P1 = ⅓P = P/3

Final temperature (T2) =?

Next, we shall convert 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273

Initial temperature (T1) = 292 K

Since the volume is constant, we can obtain the new temperature of the gas as illustrated below:

Initial pressure (P1) = P

Initial temperature (T1) = 292 K

Final pressure (P2) = P/3

Final temperature (T2) =?

P1/T1 = P2/T2

P/292 = P/3 /T2

P/292 = P/3T2

Cross multiply

P × 3T2 = 292 × P

Divide both side by P

3T2 = (292 × P)/P

3T2 = 292

Divide both side by 3

T2 = 292/3

T2 = 97.33 ≈ 97 K

Finally, we shall convert 97 K to celcius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 97

T(°C) = 97 – 273

T(°C) = –176 °C

Thus, the new temperature of the gas is –176 °C.

6 0

3 years ago

How much time is needed to deposit 1.0 g of chromium metal from an aqueous solution of crcl3 using a current of 1.5 a?

PSYCHO15rus [73]

The metal component of the given compound, CrCl3, is chromium. The number of moles per 1 g of chromium is calculated through the equation below,

n = (1 g Cr)(1 mol Cr/51.996 g Cr)
n = 0.0192 mol Cr(3 electrons/1 mol Cr)
n = 0.0577 e-

Determine the number in charge by multiplying with Faraday's constant,

C = (0.0577 mol Cr)((1 F/1 mol e-)(96485 C/ 1F)
C = 5,566.87 C

Then, calculate time by dividing the charge with the current,

t = 5566.87 C/1.5 A
t = 3711.25 minutes
t = 61.84 hours

<span><em>Answer: 61.84 hours</em></span>

8 0

3 years ago

Read 2 more answers

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