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4 years ago

What are extensive and intensive properties?

Chemistry

1 answer:

Drupady [299]4 years ago

4 0

Answer:

An extensive property is a property that depends on the amount of matter in a sample. Mass and volume are examples of extensive properties. ... Color, temperature, and solubility

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Disadvantages of Mendeleev's periodic table

mestny [16]

Answer:

Various limitations of Mendeleev's periodic table are:-

Position of hydrogen - he couldn't assign a correct position to hydrogen as it showed properties of both alkali and halogens .

Position of isotopes - he considered that the properties of elements are a function of their atomic masses. Hence isotopes of a same element couldn't be placed.

In the d-block , elements with lower atomic number were placed before higher atomic number.

Explanation:

3 0

3 years ago

Identify the products in the following chemical resction 2NaHCO3 + CaCI2 --> CaCO3 + CO2 + 2NaCI + H2O

Radda [10]

the products are:

CaCO3 Calcium Carbonate

CO2 carbon dioxide

NaCI Sodium chloride

H2O Water

8 0

3 years ago

The value of Ka for phenol (a weak acid), C6H5OH, is 1.00×10-10. Write the equation for the reaction that goes with this equilib

I am Lyosha [343]

Answer:

$Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Explanation:

Hello,

In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:

$C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+$

Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:

$K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Best regards.

3 0

3 years ago

For the reaction 2A + 4B ----> 2C + 2D, at a particular instant in time, the rate of the reaction is 0.0352 M/s. What is the

viva [34]

Explanation:

The given chemical equation is:

$2A + 4B -> 2C + 2D$

The rate of the reaction is 0.0352 M/s.

During the course of the reaction, the rate of reactants decreases, and the rate of products increases.

The rate of disappearance of B is shown below:

$rate=-\frac{1}{4} \frac{d[B]}{dt}$

So, rate of change of B is :

$rate of change of B =- rate * 4\\=-0.0352 M/s * 4\\=-0.1408M/s\\\\=-0.141M/s$

Option C.

6 0

3 years ago

The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected.

kaheart [24]

Answer:

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

7 0

3 years ago