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alexandr402 [8]

3 years ago

7

How does water move between the hydrosphere and atmosphere?

2 answers:

ratelena [41]3 years ago

7 0

Answer:

here I just googled this for you I hope it helps

Serhud [2]3 years ago

5 0

Answer:

Water moves between the hydrosphere and atmosphere through evaporation condensation and precipitation.

Explanation:

Brainliest plz

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An ethylene glycol solution contains 16.2 g of ethylene glycol (c2h6o2) in 85.4 ml of water. calculate the boiling point of the

Andre45 [30]

<span>Answer: (16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m a.) (3.9704 m) x (1.86 Â°C/m) = 7.38 Â°C change 0.00Â°C - 7.38 Â°C = - 7.38 Â°C b.) (3.9704 m) x (0.512 Â°C/m) = 2.03 Â°C change 100.00Â°C + 2.03 Â°C = 102.03 Â°C</span>

7 0

3 years ago

When a 2.00 g sample of KCl is dissolved in water in a calorimeter that has a total heat capacity of 1.28 kJ ⋅ K − 1 , the tempe

Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

$q=c\times (\Delta T)$

where,

q = heat produced = ?

c = specific heat capacity of water = $1.28kJ/K$

$\Delta T$ = change in temperature = 0.360 K

Now put all the given values in the above formula, we get:

$q=1.28kJ/K\times 0.360K$

$q=0.4608kJ=460.8J$

Now we have to calculate the molar heat solution of KCl.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 460.8 J

m = mass of $KCl$ = 2.00 g

Molar mass of $KCl$ = 74.55 g/mol

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole$

Now put all the given values in the above formula, we get:

$\Delta H=\frac{460.8J}{0.0268mole}$

$\Delta H=17194.029J/mol=17.19kJ/mol$

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

7 0

2 years ago

How good are you with chemistry

Ugo [173]

Answer:

<h2>━☆ﾟ.*･｡ﾟFine, but not as good</h2>

7 0

2 years ago

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Imagine a rock comprised entirely of an unstable isotope. After three half-lives, how much of the parent isotope remains

uysha [10]

Answer: After three half-lives 1/8 (12.5%) of the original sample remains

7 0

1 year ago

What is △n for the following equation in relating Kc to Kp?

Nimfa-mama [501]

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>

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7 0

3 years ago