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3 years ago

What is the main way that heat passes through a brick in the wall of a building?

Chemistry

1 answer:

Luden [163]3 years ago

8 0

its conduction

Hopeit helps)❤❤

Let me know if ther is an error in my answer

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The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,

Katyanochek1 [597]

Answer:

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

$n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

$n_g = 0.00215 \ mols$

Given that the number of moles of CHCl3 evaporated is less than the number of moles of CHCl3 initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of air so the pressure at equilibrium is 100 mmHg

4 0

3 years ago

HELP!!! WRITE THE CORRECT ISOTOPE NOTATION FOR EACH OF THE FOLLOW:

kakasveta [241]

The answer is 4 an atom

4 0

3 years ago

What mass of sodium bicarbonate do you start with

valentina_108 [34]

Answer:

84.007 g/mol

Explanation:

welcome

5 0

3 years ago

Why does every sample of water have the same boiling point?

Tamiku [17]

Answer:

beacuse water is same every where because it is the combination of h2 +o2 h2o which doesn't change while it is different

Explanation:

8 0

3 years ago

15.3 g of nano3 were dissolved in 100g of water in a calorimeter. The temperature of the water drops from 25.00°c to 21.56°c. Ca

Arturiano [62]

Answer:

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 100.0 g).

c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

∵ Q = m.c.ΔT

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

∵ ΔH = Q/n

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

3 0

3 years ago