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3 years ago

How many grams are there in 7.4 x 10^23 molecules of AgNO3?

Chemistry

1 answer:

mr_godi [17]3 years ago

4 0

Data we are collecting from the equation as:

7.4x10^23 = silver nitrate molecules as given from question.

As we know avogandro'S no. 6.022*1023 which is number of molecules.

So when we calculate any no. Of mole of any atom we know that 1mole of any atom is equal to avogandro'S no and grams can be calculate from moles equation.

So back to point,

Dividing molecules of AgNO3 to no. of molecules per mol of AgNO3

7.4 / 6.02*1023

= 1.229*1023per mole of AgNO3.

OK we get the moles of AgNO3

Now checking the gram of AgNO3

Weight of AgNO3 is 163.868.

Now multiplying both values moles*molar weight of AgNO3

1.229*1023 * 763.868 = 208.767grams of AgNo3

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LUCKY_DIMON [66]

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4 years ago

Choose ALL TRUE statements about calorimetry from the choices below:

Ostrovityanka [42]

Answer:

True statment

2) Styrofoam would make a good calorimeter

3) Insulating material would make a good calorimeter

Explanation:

The calorimeter is one which is insulated that is which will not absorb or let the heat to escape from it. the calorimeter is used to measure the heat change during a process so if it will allow to exchange heat with surrounding it will deviate the readings or observence.

Copper is a good conductor of heat so we cannot use it make a calorimeter.

Hence

1) Copper would make a good calorimeter : False

2) Styrofoam would make a good calorimeter: True

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3 years ago

Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h

OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q = 16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0 = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

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