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2 years ago

What are some of the unique properties of water?

Chemistry

2 answers:

Zepler [3.9K]2 years ago

4 0

Answer:

Water molecules are polar, so they form hydrogen bonds. This gives water unique properties, such as a relatively high boiling point, high specific heat, cohesion, adhesion and density

Explanation:

Vlad [161]2 years ago

4 0

It dosent have a taste and it clenches thirst

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What effect would a latitude of 60° have on the climate?

Crazy boy [7]

Answer:Temperatures Cool With Increasing Latitude

Explanation:

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2 years ago

Select the correct answer.

Lera25 [3.4K]

Answer:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$

Explanation:

S is in 3d period, so it has 3 levels

1 level has 2 electrons, 2d level 8 electrons, 3d level 6 electrons

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$

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2 years ago

A solution containing 3.90 g of an unknown nonelectrolyte liquid and 9.60 g water has a freezing point of −3.33 °C. The Kf = 1.8

pantera1 [17]

Answer:

The molar mass is 227 g/mol

Explanation:

Step 1: Data given

Mass of unknown nonelectrolyte = 3.90 grams

Mass of water = 9.60 grams

Freezing point of the solution = -3.33 °C

Kf = 1.86°C/m

Nonelectrolyte has a van't Hoff factor = 1

Step 2:

If you know the number of moles, and you know that is equivalent to 1.00 g, you can get molar mass.

∆T =i*m*K f

⇒ ∆T = difference in temperature between freezing point of solution and pure water = 3.33 °C

⇒ Van't hoff factor of the nonelectrolyte = 1

⇒ molality = moles nonelectrolyte / mass water

⇒ Kf = freezing point constant = 1.86 °C/m

3.33 = (1)(m)(1.86)

m = 1.79 molal = 1.79 moles / kg H2O

Step 3: Calculate moles nonelectrolyte

molality = moles / mass H2O

moles = molality * mass H2O

Moles = 1.79 molal * 0.0096

Moles = 0.017184‬ moles

Step 4: Calculate molar mass of nonelectrolyte

Molar mass = mass / moles

Molar mass = 3.90 grams / 0.017184 moles

Molar mass = 227 g/mol

The molar mass is 227 g/mol

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2 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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2 years ago

The gene for curled ears is dominant over the gene for straight ears (e). If you crossed a cat with curled ears (Ee) and a cat w

tatyana61 [14]

Answer:

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3 years ago