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1 day ago

A buffer is prepared by adding 100 mL of 0.50 M sodium hydroxide to 100 mL of 0.75 M propanoic acid. Is this a buffer solution,

and if so, what is its pH?

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Compounds A and B are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 g of sulf

lorasvet [2795]

Answer:

1.5

Explanation:

It is given that:

Compound A and B originate from Sulfur + Oxygen.

Compound A:

6g sulfur + 5.99g Oxygen

Compound B:

8.6g sulfur + 12.88g oxygen

By comparing the ratios:

Compound A:

S: O = 6.00: 5.99

S/0 = 6.0g S / 5.99g O

Compound B:

S: O = 8.60: 12.88

S / O = 8.60g S / 12.88g O

The mass ratio of A and that of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

= 77.28 / 51.514

= 1.50017

= 1.5

4 0

2 months ago

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

lorasvet [2795]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Let's examine the following serial dilution processes.

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is further diluted to 250 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM:

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM:

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is further diluted to 500 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

7 0

2 months ago

Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO.

lions [2927]

The empirical formula is K₂CO₃.

This formula represents the most simplified whole-number ratio of atoms in a chemical compound.

The atom ratio aligns with the mole ratio, which means our task is to find the molar ratios for K:C:O.

I prefer to summarize these calculations in table form.

Element Moles Ratio¹ Integers²

K 0.104 2.00 2

C 0.052 1.00 1

O 0.156 3.00 3

¹ To obtain the molar ratio, each mole value is divided by the smallest mole count.

² Convert the ratio values to integers (2, 1, and 3).

The empirical formula is K₂CO₃.

6 0

2 months ago

Which of the following pairs lists a substance that can neutralize Ca(OH)2 and the salt that would be produced from the reaction

Anarel [2989]

Answer:

HCl and CaCl2

Explanation:

Calcium hydroxide acts as a base, thus requiring an acid for neutralization.

HCl functions as an acid, which can neutralize calcium hydroxide when they react together.

The resulting salt will be calcium chloride (CaCl2).

The interaction between calcium hydroxide and HCl yields Ca(OH)Cl and CaCl2.

Here, Ca(OH)Cl exists as a solution and calcium chloride serves as a base.

I hope this response assists you.

7 0

2 months ago

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