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3 years ago

What is enthalpy of atomisation

1 answer:

5 0

The enthalpy of atomization (also atomisation in British spelling) is the enthalpy change that accompanies the total separation of all atoms in a chemical substance (either a chemical element or a chemical compound)

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Crude oil is ________. a. composed of less than ten different hydrocarbon molecules b. formed in a wide range of temperature and

Arisa [49]

Answer:

Explanation:

A - Crude oil is composed for hundreds of hydrocarbon, not less than ten.

B - Is formed in specific conditions of temperature and pressure

C - It's fractionated to form gasoline, lubricants, CH4, plastics and many other products made of hydrocarbon.

D - We have crude oil located more than 4000 yards below the surface in countries like Brazil

E - The crude oil is very thick and don't have an specific usage, so we need to refine it.

3 0

3 years ago

Jimmy's Chemical reaction/ Recipe for making Grilled chese sandwiches is given below: 1 slice of cheese + 2 slices of bread = 1

Juli2301 [7.4K]

Answer:

See explanation

Explanation:

We have been told in the question that the equation of the reaction is; 1 slice of cheese + 2 slices of bread = 1 Grilled cheese sandwich ( mole ratio is, 1:2:1) .

Then the reagents are 10 slices of cheese 30 slices of bread. It then follows that 10 slices of cheese should be combined with 20 slices of bread according to the mole ratio.

However, we have 30 slices of bread and 10 slices of cheese so cheese is the limiting reactant while bread is the reactant in excess.

Yes, the number of glilled chese sandwishes he can make is decided by the limiting reactant because it gets used up most.

4 0

3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

4 years ago

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$