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Helen [10]
3 years ago
11

Which of the following aqueous solutions are good buffer systems?

Chemistry
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

0.25 M ammonium nitrate + 0.40 M ammonia  

0.39 M hypochlorous acid + 0.25 M potassium hypochlorite

0.12 M hydrofluoric acid + 0.14 M sodium fluoride  

Explanation:

A buffer consists of a weak acid and its salt or a weak base and its salt.

NH₃ is a weak base, and HClO and HF are weak acids.

B is wrong. HNO₃ is a strong acid.

C is wrong. KOH is a strong base.

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Through a process known as electrolysis, water can be split into hydrogen and oxygen. What type of reaction is this? Give an exp
AlexFokin [52]
Electrolysis of water<span> is the </span><span>decomposition reaction, because from one molecule (water) two molecules (hydrogen and oxygen) are produced. Water is separeted into two molecules:
</span>Reaction of reduction at cathode: 2H⁺(aq) + 2e⁻<span> → H</span>₂(g<span>).
</span><span><span>Reaction of oxidation at anode: 2H</span></span>₂<span><span>O(l) → O</span></span>₂<span><span>(g) + 4H</span></span>⁺(<span><span>aq) + 4e</span></span>⁻.<span><span>
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8 0
3 years ago
In which situation would hydrogen bonding be present?​
sineoko [7]

Answer:

Water ?

Need more info.

Water is H_{2}<em>O</em>

Explanation:

6 0
3 years ago
Read 2 more answers
A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
deff fn [24]

Answer:

A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

Explanation:

From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

            A = Pre exponetial factor

            E_{a} = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

T_{1}=505K

T_{2}=525K

R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
3 years ago
The hot dogs you ate at the barbecue last week were 75% fat-free by weight, had 275 calories, and weighed 110 g. a) What percent
vovangra [49]

Answer:

a) 25%

b) 27.5 g

c) 90%

Explanation:

a) 75% fat-free by weight means 25% of the weight is made by fat.

b) 110 g ___ 100%

      x    ___ 25%

          x = 27.5g

Each hot dog has 27.5g of fat.

c) 9 cal ___ 1 g fat

      y    ___ 27.5 g fat

          y = 247.5 cal

275 cal ___ 100%

247.5 cal ___ z

     z = 90%

90 % of the calories come from fat.

6 0
3 years ago
Help!
babunello [35]
D.mno4- is reuced it loses h atom

7 0
3 years ago
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