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Reika [66]
3 years ago
9

What are 5 intensive physical properties of water?

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
3 0
Given the low molar mass of its constituent molecules, water has unusually large values ​​of viscosity, surface tension, heat of vaporization, and entropy of vaporization, all of which can be ascribed to the extensive hydrogen bonding interactions present in liquid water.
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Determine the hydrogen ion concentration of solution B. [1]
bazaltina [42]
The hydrogen Ion concentration of solution  B is 
1.0 x  10^-5 or 0.000 010 M

You can see that this will be proportional to the amount of B's PH compared to A's

hope this helps
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3 years ago
Can someone help me please I’m giving the brainliest
Len [333]
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3 years ago
What genotype will appear in box 4?
Pani-rosa [81]
The genotype will be (tt)
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7 0
3 years ago
Read 2 more answers
A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0g sample of a pure substance from to −
astraxan [27]

Answer:

C = 0.2349 J/ (g °C)

Explanation:

Mass, m = 894.0g

Initial Temperature = −5.8°C

Final Temperature =  17.5°C

Temperature change = 17.5°C - (−5.8°C) = 23.3

Heat, H = 4.90kJ = 4900 J

Specific heat capacit, C = ?

The relationship between these quantities is given by the equation;

H = mCΔT

C = H / mΔT

C = 4900 / (894)(23.3)

C = 0.2349 J/ (g °C)

4 0
3 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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