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Ostrovityanka [42]
3 years ago
10

How much heat (in Joules) is needed to raise the temperature of 257g of ethanol (cethanol=2.4 J/g°C) by 49.1°C?

Chemistry
1 answer:
Sonbull [250]3 years ago
8 0

Answer:

Q = 30284.88 j

Explanation:

Given data:

Mass of ethanol = 257 g

Cp = 2.4 j/g.°C

Chnage in temperature = ΔT = 49.1°C

Heat required = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = 257 g× 2.4 j/g.°C × 49.1 °C

Q = 30284.88 j

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The compound methylamine, CH3NH2, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibrium
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Answer:

Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]

Explanation:

According to Brönsted-Lowry acid-base theory:

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  • A base is a substance that accepts H⁺.

When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.

CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)

The basic equilibrium constant (Kb) is:

Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]

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3 years ago
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3 years ago
At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepare
OleMash [197]

Answer:

P_{CCl_4}=52.43torr

Explanation:

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In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

P_{CCl_4}=x_{CCl_4}P_{CCl_4}^{vap}

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

x_{CCl_4}=\frac{140/153.81}{140/153.81+67.1/100.21}=0.576

Therefore, the partial pressure of chloroform turns out to be:

P_{CCl_4}=0.576*91torr\\\\P_{CCl_4}=52.43torr

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3 0
3 years ago
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
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