Question 22

a 22.44g sample of iron absorbs 180.8 J of heat, upon which the temperature of the sample increases from 21.1C to 39.0/ what is

NeX [460]

Answer:

The specific heat of iron is 0.45 J/g.°C

Explanation:

The amount of heat absorbed by the metal is given by:

heat = m x Sh x ΔT

From the data, we have:

heat = 180.8 J

mass = m = 22.44 g

ΔT = Final temperature - Initial temperature = 39.0°C - 21.1 °C = 17.9°C

Thus, we calculate the specific heat of iron (Sh) as follows:

Sh = heat/(m x ΔT) = (180.8 J)/(22.44 g x 17.9°C) = 0.45 J/g.°C

8 0

3 years ago

Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g

Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

$\frac{Mass}{Volume}$

$Mass = \frac{percent \;of \;ethanol \;X \;density \;of \;ethanol \;+ \\ \;percent \;of \;water X \;density \;of \;water}{100}$

$0.926 = \frac{x X 0.7890 g/mL \;+ (100-x) X 0.9982 g/mL}{100}$

$x= 34.51$ %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

Learn more about the density here:

brainly.com/question/952755

#SPJ1

8 0

1 year ago

____________ properties include melting point, boiling point, strength, and malleability.a.physicalc.chemicalb.reactived.none of

lina2011 [118]

Physical properties are those which can be observed without any change in composition of the substance. Hence, a is the answer.

8 0

3 years ago

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$

Molar mass of A = $M_A$

The depression in freezing point of solution with B solute: $\Delta T_{f,B}$

Molar mass of B = $M_B$

$\Delta T_{f,A}>\Delta T_{f,B}$

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

$\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}$

$M_A$

This means compound B has greater molar mass than compound A,

4 0

3 years ago

When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample

sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

<u>Step 2:</u> Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid * 26.38 kJ/g = 6.4631 kJ

<u>Step 3</u>: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ / 1.643°C = 3.934 kJ/ °C

<u>Step 4:</u> Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine

5.65 kJ / 0.0013389 moles = 4219.9 kJ/mol ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0

3 years ago