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3 years ago

Calculate the heat required to melt 25.7 g of solid methanol at its melting point

Chemistry

2 answers:

gulaghasi [49]3 years ago

7 0

Answer:

the heat required to melt the compound is 2.56k

umka2103 [35]3 years ago

4 0

The heat required to melt the compound is 2.56kJ.

Why?

To calculate the heat required to melt 25.7 g of solid methanol, we need to calculate its molar mass, know its chemical formula, its molar heat of fusion, and calculate how many moles represent 25.7g of the same compound.

Then, we need to use the following formula to calculate the heat:

$Q=N*u$

Where,

N is the moles of the compound

u, is the molar heat of fusion of the compound (J/mol)

We know that the molar mass of methanol is :

$32.042\frac{g}{mol}$

So, we can calculate the number of moles that represents 25.7g of the same compound:

$N=\frac{mass(g)}{molarmass(\frac{g}{mol}} \\\\N=\frac{25.7g}{32.042\frac{g}{mol}}=0.8mol(methanol)$

Also, the molar heat of fusion of methanol is:

$u_{methanol}=3200\frac{J}{mol}=3.2\frac{kJ}{mol}$

Now, substituting into the first equation and calculating, we have:

$Q=N*u$

$Q=0.8mol*3.2\frac{kJ}{mol}=2.56kJ$

Hence, we have that the heat required to melt the compound is 2.56kJ.

Have a nice day!

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Write the chemical formula of tetraphosphours octasulfide

ArbitrLikvidat [17]

Answer:

P4 S8

Explanation:

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3 years ago

The colligative molality of an unknown aqueous solution is 1.56 m.

yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- Vapor pressure lowering

ΔP = P° . Xm . i

- Boiling point elevation

ΔT = Kb . m . i

- Freezing point depressión

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

3 0

4 years ago

What type of reaction is 3Ca(OH)2 + Al2(SO4)3 = 3CaSO4 + 2Al(OH)3 Combustion, Synthesis, Decomposition, Single Displacement, D

Alexeev081 [22]

Answer is: Double Displacement.
Combustion is reaction with oxygen.
Synthesis is reaction of two or more substances combining to make a more complex substance.
Decomposition is reaction where one substance is broken down into two or more simpler substances.
Single Displacement is reaction where neutral element metal or nonmetal become an ion as it replaces another ion in a compound.
Double displacement reactions (more reactive metals displace metals with lower reactivity).
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3 0

3 years ago

In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o

Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

$V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1} )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w$

The specific gravity of the wood chips is 0.494.

The average volume of a log is

$V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2} \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}$

The weight of one log is

$M=\rho*V=0.494*\rho_w*12.57 ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm$

To provide 3617 ton/day of wood chips, we need

$n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min$

The feed rate of logs is 14.3 logs/min.

7 0

3 years ago

Read 2 more answers

A sample of gas occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 at

alina1380 [7]

P₁ = 0.90 atm

V₁ = 50.0 mL

T₁ = 298 K

P₂ = 1 atm

T₂ = 273 K

V₂ = P₁ x V₁ x T₂ / T₁ x P₂

V₂ = 0.90 x 50.0 x 273 / 298 x 1

V₂ = 12285 / 298

V₂ = 41 mL

Answer (1)

hope this helps!

7 0

3 years ago