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3 years ago

How did lavoisier revolutionize the science in chemistry

1 answer:

3 0

Lavoisier revolutionised chemistry science with his discovery of the role oxygen plays in combustion. He recognised and labelled the elements oxygen and hydrogen; he was also largely opposed to the phlogiston theory which dictates that a fire-like element called phlogiston is contained within combustible elements and is released during combustion. These were his most noted contributions to science.

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A flexible container holds 32.4L of gas at 55.0c. What will be the volume if the temperature falls to 17.0c

mash [69]

The answer will be 25.92 L for 17.0 c

5 0

3 years ago

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

If copper carbonate system is heated for too short a time, what will you notice for mass percent of copper be too large or too s

sammy [17]

Answer:

The mass percent of copper as element is the same.

Explanation:

First of all we need the reaction that is presented below:

$CuCO_{3 (s)} +Heat$ → $CuO_{(s)} + CO_{2 (g)}$

The mass percent of copper (Cu) as element is the same because of during the reaction the element only transform its nature from copper carbonate ( $CuCO_{3 (s)}$ ) to copper oxide ( $CuO_{(s)}$ ), the latter is a solid and will remain in the system.

On the other hand, you will note that the global percentage mass will be small because of the reaction produce ( $CO_{2 (g)}$ ) that is a gas and this one will escape for the system.

Have a great day!

3 0

3 years ago

Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume wou

lesya [120]

First, let's compute the number of moles in the system assuming ideal gas behavior.

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles

At standard conditions, the standard molar volume is 22.4 L/mol. Thus,

Standard volume = 22.4 L/mol * 2.176 mol = 48.74 L

3 0

3 years ago

How much heat does a 100. g sample of copper absorb when its temperature increases by 30.0°C? The specific heat of copper is 0.3

erica [24]

Answer:

$\boxed {\boxed {\sf B. \ 1170 \ Joules }}$

Explanation:

We are asked to find how much heat a sample of copper absorbs when the temperature is increased.

Since we know the mass, temperature increase, and specific heat capacity, we can use the following formula to calculate heat.

$q= mc \Delta T$

The mass of the copper sample is 100 grams, the temperature is changed or increased by 30.0 degrees Celsius, and the specific heat of copper is 0.39 Joules per gram degrees Celsius.

m= 100 g
c= 0.39 J/g °C
ΔT= 30.0 °C

Substitute the values into the formula.

$q= (100 \ g )(0.39 \ J/g \textdegree C ) (30.0 \textdegree C )$

Multiply the first two values. Note that the units of grams cancel.

$q= 39 \ J/ \textdegree C (30.0 \textdegree C )$

Multiply again, this time the units of degrees Celsius cancel.

$q= 1170 \ J$

The copper sample absorbs 1170 Joules of heat and Choice B is correct.

7 0

3 years ago