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creativ13 [48]
3 years ago
13

A student carried heated a 25.00 g piece of aluminum to a temperature of 100°C, and placed it in 100.00 g of water, initially at

a temperature of 10.0°C. Determine the final temperature of the system (aluminum and water). Useful data: The specific heat of water is 4.18 J/(g°C) The specific heat of aluminum is 0.900 J/(g°C)
Chemistry
1 answer:
SashulF [63]3 years ago
4 0

Answer:

The final temperature of the system is 14.6 °C

Explanation:

<u>Step 1:</u> Data given

mass of the aluminium = 25.00 grams

mass of the water = 100.00 grams

Initial temperature of aluminium = 100 °C

Initial temperature of water = 10.0 °C

Specific heat of water = 4.18 J/g°C

Specific heat of aluminium 0.900 J/g°C

<u>Step 2:</u> Heat transfer

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(aluminium) * C(aluminium) * ΔT(aluminium) = - m(water) * C(water) * ΔT(water)

25*0.900*(T2-100) = - 100*4.18 * ( T2 - 10)

2250 - 22.5T2 = 418T2 - 4180

T2 =14.6 °C

The final temperature of the system is 14.6 °C

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2 years ago
How many miles of carbon dioxide gas can be produced if 25 g of cr2(co3)3 are reacted with excess nitric acid hno3 ?
Mars2501 [29]

Answer:

0.264 moles of carbon dioxide are produced.

Explanation:

<u>Limiting reagent</u> : The reactant that is consumed in a reaction is called the limiting reagent.

Molar mass = mass of substance  in grams present in 1 mole of the compound.

before solving this question,

It is needed to know , the molar mass of  cr2(co3)3 and CO2

Molar mass of  cr2(co3)3  = 2(mass of Cr) + 3(massof C) + 9(mass of O)

                                           = 2(52) + 3(12) + 9(16)

                                          = 104 + 36 + 144

                                          = 284 g

This means 1 mole of  cr2(co3)3  = 284 g............(1)

Molar mass of CO2 = mass of C + 2(mass of O)

                                  = 12 + 2(16)

                                  = 12 + 32

                                  = 44 g

This means 1 mole of CO2 contain 44 g of it.

So, 3 mole of CO2 contain = 44x3 g

3 mole of CO2 contain = 132 g.............(2)

The balance equation for the reaction is :

Cr_{2}(CO_{3})_{3}+6HNO_{3}\rightarrow

2Cr(NO_{3})_{3} +8H_{2}O+3CO_{2}

HNO3 = excess reagent , it can't be used in for predicting the moles of CO2

Cr_{2}(CO_{3})_{3}= Limiting reagent = It will tell the amount of CO2 that will be produced

In the balanced equation ,

1 mole of cr2(co3)3  is giving = 3 mole of CO2

Use equation (1) and (2)

284 g of cr2(co3)3  is giving = 132 g of CO2

1 gram of cr2(co3)3  is giving =

\frac{132}{284} g of CO2

25 g of cr2(co3)3 will give =

\frac{132}{284}\times 25 g of CO2

= 11.62 gram CO2

now , moles of CO2 can be calculated using ,

moles=\frac{given\ mass}{Molar\ mass\ of\ CO2}

moles =\frac{11.6}{44}      (molar mass of co2 = 44 g)

moles of CO2 = 0.264

6 0
3 years ago
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.4-L bulb, then filled
vazorg [7]

Answer : The diatomic gas is nitrogen gas, N₂.

Explanation :

First we have to calculate the moles of gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of gas = 1.00 atm

V = Volume of gas = 4.4 L

n = number of moles of gas = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 22.0^oC=273+22.0=295.0K

Putting values in above equation, we get:

1.00atm\times 4.4L=n\times (0.0821L.atm/mol.K)\times 295.0K

n=0.1817mol

Now we have to calculate the molar mass of gas.

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As we are given that the gas is diatomic X₂.

As, 2 atoms of gas X has mass = 28.07 g/mol

So, 1 atom of gas will have mass = \frac{28.07}{2}=14.04g/mol

From this we conclude that the nitrogen atom has mass of 14.04 g/mol.

Thus, the diatomic gas is nitrogen gas, N₂.

7 0
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