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3 years ago

Pls help Plot the point (5,9)

Chemistry

2 answers:

Varvara68 [4.7K]3 years ago

6 0

Go five across the x axis, then go nine straight up from there

elena-s [515]3 years ago

4 0

All u do is go to 5 and up nine and just put a little cross

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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

How many grams of iron can be obtained from a 268g sample of iron(III) oxide?

Nikolay [14]

Answer:187

Explanation:

5 0

4 years ago

How many periods does the modern periodic table have?

dusya [7]

There are 7 periods in the modern periodic table

8 0

3 years ago

A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration,

Aleksandr-060686 [28]

Answer:

percentage of water = 18.76%

Explanation:

A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams.

The overall weight of the compound is 0.4813 grams.

weight of hydrated sample = 0.4813 grams.

weight of anhydrous compound = 0.391 grams

percentage weight of water = mass of water/mass of the hydrated compound × 100

mass of water = mass of hydrated compound - mass of anhydrous compound

mass of water = 0.4813 - 0.391 = 0.0903 grams.

percentage of water = 0.0903/0.4813 × 100

percentage of water = 9.03/0.4813

percentage of water = 18.761687097

percentage of water = 18.76%

6 0

4 years ago

Read 2 more answers

Help help help !!!!!!

11111nata11111 [884]

Answer:

3. 116.5 V

4. 119.6 V

Explanation:

3. Determination of the voltage.

Resistance (R) = 25 Ω

Current (I) = 4.66 A

Voltage (V) =?

V = IR

V = 4.66 × 25

V = 116.5 V

Thus, the voltage is 116.5 V

4. Determination of the voltage.

Current (I) = 9.80 A

Resistance (R) = 12.2 Ω

Voltage (V) =?

V = IR

V = 9.80 × 12.2

V = 119.6 V

Thus, the voltage is 119.6 V

4 0

3 years ago