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RSB [31]
2 years ago
15

What is the minimum mass of 25°c water required to melt all 35.1 g of ice?

Chemistry
1 answer:
inn [45]2 years ago
5 0
ΔH fusion of water is 6.0 kJ/mol
mass of ice = 35.1 g
Molecular weight of water = 18.015 g / mole
number of moles of water = 35.1 / 18.015 = 1.95 mole
amount of heat required to melt ice = moles * Δ H fusion = 1.95 x 6 =11.7 kJ
mass of water at 25°C required to get this energy to a temperature 0°C
q = m C ΔT 
11700 J = m * 4.184 * (25°- 0°)
m = 111.85 g  

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If the temperature of a gas is increased from 20°C to 35°C, what is the new pressure if the original pressure was 1.2 atm? Assum
Marta_Voda [28]

Answer:

B.) 1.3 atm

Explanation:

To find the new pressure, you need to use Gay-Lussac's Law:

P₁ / T₁ = P₂ / T₂

In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.

P₁ = 1.2 atm                                    P₂ = ? atm

T₁ = 20 °C + 273 = 293 K              T₂ = 35 °C + 273 = 308 K

P₁ / T₁ = P₂ / T₂                                             <----- Gay-Lussac's Law

(1.2 atm) / (293 K) = P₂ / (308 K)                  <----- Insert values

0.0041 = P₂ / (308 K)                                   <----- Simplify left side

1.3 = P₂                                                         <----- Multiply both sides by 308

3 0
1 year ago
What atom is the smallest size? A. Li B. Be C. F D. C
s2008m [1.1K]

Answer:

A

Explanation:

lies to the left of periodic table

7 0
3 years ago
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

5 0
3 years ago
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