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Hunter-Best [27]
2 years ago
7

You have been given two samples of water. One is from the atlantic ocean and the other is from lake superior. Describe how you w

ill tell them apart without tasting them?
Chemistry
1 answer:
lbvjy [14]2 years ago
5 0

Answer:

i think by their salinity (the saltiness or amount of salt dissolved in a body of water)

Explanation:

Water salinity is measured by passing an electric current between the two electrodes of a salinity meter in a sample of water.

You might be interested in
Which of the following would be a reasonable synthesis of CH3CH2CH2CH2OH?Group of answer choices
vampirchik [111]

Answer:

B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-

Explanation:

In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.

the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.

8 0
3 years ago
The gasses in a hair spray can are at temperature 300k and a pressure of 30 atm, it
Sergeeva-Olga [200]

Answer:

900 K

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

Because only pressure and temperature is changing, we can rearrange the equation as follows:
\displaystyle \frac{P}{T} = \frac{nR}{V}

The right-hand side stays constant. Therefore:

\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}

The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.

Substitute and solve for <em>T</em>₂:

\displaystyle \begin{aligned} \frac{(30\text{ atm})}{(300\text{ K})} & = \frac{(90\text{ atm})}{T_2} \\ \\ T_2 & = 900\text{ K}\end{aligned}

Hence, the temperature must be reach 900 K.

7 0
2 years ago
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
3 years ago
Identify the conjugate acid/base pairs in each of the following equations:
Valentin [98]

Answer:

(a) Pair 1: H₂S and HS⁻

    Pair 2: NH₃ and NH₄⁺

(b) Pair 1: HSO₄⁻ and SO₄⁻

    Pair 2: NH₃ and NH₄⁺

(c) Pair 1: HBr and Br⁻

    Pair 2: CH₃O⁻ and CH₃OH

(d)  Pair 1: HNO₃ and NO₃⁻

     Pair 2: H₃O⁺

Explanation:

When an acid loses its proton (H⁺), a conjugate base is produced.

When a base accepts a proton (H⁺), it forms a conjugate acid.

(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.

    NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺

(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.

     The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.

(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.

   CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.

(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.

    H₂O gains a proton to form the conjugate acid H₃O⁺.

6 0
3 years ago
What is an ionic bond in an electron?
Igoryamba
Its the complete transfer of valence electrons between atoms 
7 0
3 years ago
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