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melamori03 [73]
3 years ago
12

Calculate the number of atoms in 4.2 moles of sulfur atoms.

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
7 0
1 mole = 6.022*10^23 atoms
4.2 moles sulfur = 4.2* 6.022 * 10^23 sulfur atoms = 2.421*10^24 sulfur atoms

1 mole sulfur = <span>32.065 g sulfur
4.2 moles sulfur = 4.2 * </span>32.065 g sulfur = 128.901 g sulfur
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How would the pH value of an aqueous solution change, when the hydronium ion concentration is increased by a factor of 10?
Anna71 [15]

Answer:

Increasing H⁺ by 10x => pH decreases by 1 unit

Explanation:

In general, adding H⁺ ions to any aqueous solution ALWAYS causes pH values to fall ( decrease ). Just as adding OH⁻ ions to an aqueous solution causes pH values to rise ( increase ).

Here's a simple calculation demonstrating this...

Given 0.01M HCl(aq) => 0.01M H⁺(aq) + Cl⁻(aq) => pH = -log(0.01) = 2.00

Increase [H⁺] by 10x => 0.10M H⁺(aq) => pH = -log[H⁺] = -log(0.10) = 1.00

Solution with higher H⁺ concentration shows <u>pH decreasing by 1 unit.</u>

______________________________________________________-

Just to support the above statement about adding OH⁻ ions showing an increase in pH values, the following is also provided FYI ..

Given 0.01M NaOH(aq) => 0.01M OH⁻(aq) + Na⁺(aq) => pOH = -log(0.01) = 2.00 => pH = 14 - pOH = 14 - 2 = 12

Increase [OH⁻] by 10x => 0.10M OH⁻(aq) => pOH = -log[OH⁻] = -log(0.10) = 1.00 => pH = 14 - pOH = 14 - 1 = 13

Increasing [OH⁻] by 10x => <u>increasing pH by 1 unit. </u>

Solution with higher H⁺ concentration shows pH decreasing by 1 unit.

______________________________________________________

Remember, for <u>any</u> aqueous solution ...

=> Adding H⁺   => always decreases pH

=> Adding OH⁻ => always increases pH

4 0
3 years ago
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otez555 [7]

Answer:

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Explanation:

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6 0
2 years ago
CH3 + HCl &lt;=&gt; CH3Cl + H2O
dmitriy555 [2]

Answer:

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

Explanation:

Step 1: Data given

Kp = 4.7 x 10^3 at 400K

Pressure of CH3OH = 0.250 atm

Pressure of HCl = 0.600 atm

Volume = 10.00 L

Step 2: The balanced equation

CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)

Step 3: The initial pressure

p(CH3OH) = 0.250atm

p(HCl) = 0.600 atm

p(CH3Cl)= 0 atm

p(H2O) = 0 atm

Step 3: Calculate the pressure at the equilibrium

p(CH3OH) = 0.250 - X atm

p(HCl) = 0.600 - X atm

p(CH3Cl)= X atm

p(H2O) = X atm

Step 4: Calculate Kp

Kp = (pHO * pCH3Cl) / (pCH3* pHCl)

4.7 * 10³ =  X² /(0.250-X)(0.600-X)

X = 0.249962

p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm

p(HCl) = 0.600 - 0.249962 = 0.350038 atm

p(CH3Cl)= 0.249962 atm

p(H2O) = 0.249962 atm

Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)

Kp = 4.7 *10³

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

4 0
2 years ago
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Serhud [2]

Explanation:

the answer is that prokaryotic cells are not multicellular

6 0
2 years ago
How does evidence of chemical
Sliva [168]

Answer:

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Explanation:

4 0
2 years ago
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