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GuDViN [60]
4 years ago
11

Why does hard water taste better than distilled water?

Chemistry
1 answer:
Maurinko [17]4 years ago
8 0

Answer:

because hard water is salty

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Calculate the [H3O+] for orange juice, which has a [OH–] = 5.0 × 10–11 M.
zloy xaker [14]

Answer:

i Think.... i think its a carrot

Explanation: 5.0 x 10-11 M

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3 years ago
A liquid dissolved in a gas results in a solution that is _____.
SVETLANKA909090 [29]

The correct answer is C


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The maximum efficiency possible in an energy conversion process that is not limited by the second law of thermodynamics
Anna [14]

Answer:

100 %

Explanation:

The maximun efficiency possible (whem not limited by the  second law of thermodynamics) happens when all the energy used is transformed into the type of energy we required with no other transformations.

For example, in an engine we want that all the energy we supply is being converted to work. That's the ideal case, but in reality always some of that energy is lost in the form of heat.

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3 years ago
Explain why the C―Ha bond is much more acidic than the C―Hb bond in pentan-2-one. Select the single best answer. 253 Ha is less
mafiozo [28]

Answer:

Ha is more acidic than Hb because loss of Ha forms a resonance-stabilized conjugate base.

Explanation:

The carbon atom that is next to the carbonyl group in pentan-2-one is known as the alpha carbon atom, this carbon atom bears the Ha, the alpha hydrogen atoms.

Ha is more acidic than Hb because, loss of Ha leads to the formation of a resonance stabilized enolate ion. This resonance stabilization of the ion formed makes loss of Ha an easier process than loss of Hb, hence the answer above.

3 0
4 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
3 years ago
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