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2 years ago

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

Chemistry

1 answer:

NemiM [27]2 years ago

3 0

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

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According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that ther

IgorC [24]

Answer:

0.020 moles of $Fe(OH)_{2}$ can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

$M=\frac{molesofsolute}{Litersofsolution}$

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

$175.0mL*\frac{1L}{1000mL}=0.175L$

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of $Fe(OH)_{2}$ produced:

$FeCl_{2}(aq)+2LiOH(aq)=Fe(OH)_{2}(s)+2LiCl(aq)$

As the problem says that there are excess of $FeCl_{2}$ , the limiting reagent is the LiOH.

$0.040molesLiOH*\frac{1molFe(OH)_{2}}{2molesLiOH}=0.020molesFe(OH)_{2}$ can be formed

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2 years ago

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vladimir2022 [97]

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2 years ago

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A chunk of dry ice, solid CO2, "disappears" after sitting at room temperature for a while. There is no puddle of liquid. What ha

Vika [28.1K]

I believe the answer is A.

8 0

2 years ago

A 2.50 g sample of zinc is heated, and then placed in a calorimeter containing 65.0 g of water. Temperature of water increases f

swat32

718.65 degrees is the initial temperature of the zinc metal sample.

Explanation:

Data given:

mass of zinc sample = 2.50 grams

mass of water = 65 grams

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J

q zinc = 2.50 x 0.390 x (22.50- Ti)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.

6 0

2 years ago

What is the volume, in liters, occupied by 1.73 moles of N2 gas at 0.992 atm pressure and a temperature of 75º C? (R value- 0.08

taurus [48]

Volume of the nitrogen gas = 49.8 L

Explanation:

It is given that the pressure, number of moles and temperature of nitrogen gas, and gas constant value being constant and it is taken as 0.08206 L atm mol⁻¹K⁻¹.

Temperature = T = 75°C = 75 + 273 = 348 K

Pressure = P = 0.992 atm

Number of moles = n = 1.73 moles

We have to use the ideal gas equation, PV = nRT, and rearranging the equation to get Volume in litres.

V = $$\frac{nRT}{P}$

= $$\frac{1.73\times 0.08206\times348}{0.992}$

= 49.8 L

So the volume of Nitrogen gas = 49.8 L

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1 year ago