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3 years ago

11

Which of the following aqueous solutions are good buffer systems?

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

babunello [35]

D.mno4- is reuced it loses h atom

7 0

3 years ago