The first particles to evaporate from a liquid are those

Is distilled water homogeneous or heterogeneous?

Rama09 [41]

It is neither, it is a pure substance.

4 0

3 years ago

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an

professor190 [17]

Answer:

1.717 J/g °C ( third option)

Explanation:

A piece of the unknown metal dropped into water this means that Q of metal is equal to Q of the water. We write this equality as follows:

Step 1: writing the formulas:

Q = mc∆T

⇒ -Q(metal) = Q(water) Because :Metal dropped into water this means that Q of metal is equal to Q of the water.

We can write the formula different :

Mass of metal * (cmetal)(ΔT) = Mass of water *(cwater) (ΔT)

⇒ Here c is the specific heat and depends on material and phase

For this case :

mass of the metal = 68.6g

mass of the water = 42g

Specific heat of the metal = TO BE DETERMINED

Specific heat of the water = 4.184J/g °C

Initial temperature of the metal = 100 °C ⇒ Change of temperature: 52.1 - 100

Initial temperature of the water = 20°C ⇒ Change of temperature: 52.1 - 20

Step 2: Calculating specific heat of the metal

-(Mass of metal) * (cmetal)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-68.6g (cmetal)(52.1 - 100) = 42g (4.184j/g °C) (52.1 - 20)

-68.6g *cmetal * (-47.9) = 42g (4.184j/g °C) *(32.1)

3285.94 * cmetal = 5640.87

cmetal = 5640.87 / 3285.94 = 1,71667 J/g °C

cMetal = 1.717 J/g °C

4 0

3 years ago

PLEASE HELP HELP I NEED THIS ASAP

Sergeu [11.5K]

Wait SORRY HELP WITH WHAT?!!

5 0

3 years ago

Read 2 more answers

Which of the following are drugs that are important sources of toxicity because of their wide availability, otc status, and unin

Tom [10]

The correct answer is A. Analgesic are used to treat all forms of pain and because of their wide availability people buy them without proscription. This leads to a lot of undesirable side effects.

8 0

3 years ago

Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe

Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 $\frac{J}{g*C}$ and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 $\frac{J}{g*C}$ *20 C and solving: Q=420 J

Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g* $6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}$ = 3.333 kJ= 3,333 J (being kJ=1,000 J)

Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 $\frac{J}{g*C}$ and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 $\frac{J}{g*C}$ *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

3 0

3 years ago