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oksian1 [2.3K]
2 years ago
15

What are examples of models

Chemistry
1 answer:
a_sh-v [17]2 years ago
6 0
Here are 3 main types of models are ( Physical, Mathematical, and Conceptual. Models have limitations but are useful and can be changed based on a new evidence.
You might be interested in
What mass of sulfur would have precisely 4.7x10^22 atoms of sulfur?
Svetllana [295]

Answer:

2.498g

Explanation:

First off it's important to know Avogadro's number relationship with number of moles and molar mass.

It is given as;

1 mole = 6.02 * 10^23 atoms.

A mole's weight is equal to it's molar mass. Hence the equation can be changed to;

1 mole = molar mass = 6.02 * 10^23 atoms

In this problem, molar mass of Sulphur is 32g/mol. This means 1 mole of sulphur weighs 32 g and contains 6.02 * 10^23 atoms. What mass of sulphur would then contain 4.7x10^22 of atoms?

This leads us to;

32 g =  6.02 * 10^23

x g =  4.7x10^22

Upon cross multiplication, we have;

x =  (4.7x10^22 * 32) / 6.02 * 10^23

x = 24.98 * 10^-1

x = 2.498g

7 0
3 years ago
Read 2 more answers
Question 3 of 10
yulyashka [42]

Answer: A.) Look at the benefits and challenges of using specific materials.

Explanation: They can’t decide whether they made the best product or not because it is a new project.

6 0
1 year ago
What is the empirical formula of a compound that contains 6.10 g of hydrogen and 28 g of nitrogen?
Viktor [21]

Answer:

NH₃

Explanation:

mass H = 6.10 grams

mass N = 28.00 grams

mass cpd = (6.10 + 28.00)grams = 34.10 grams

%H/100wt = (6.10/34.10)100% = 17.9% w/w

%N/100wt = (28.00/34.1)100% = 82.1% w/w

%/100wt => grams/100wt => moles => ratio => reduce => emp ratio

%H/100wt = 17.9% w/w => 17.9g => (17.9/1)moles = 17.9 moles H

%N/100wt = 82.1% w/w => 82.1g => (82.1/14)moles = 5.9 moles N

Ratio N:H => 17.9 : 5.9

Reduce mole ratio (divide by smaller mole value) => 17.9/5.9 : 5.9/5.9

=> 3HY:1H empirical ratio => empirical formula NH₃ (ammonia)

3 0
3 years ago
An isotope of potassium with a half-life of roughly one billion years is often used in radioactive dating. This isotope decays t
melisa1 [442]

Answer : The chemical symbol for the element that results from this process is, (Ar) for argon.

Explanation :

Electron capture : In this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. Simply, a proton and an electron combines together to form a neutron. Mass number does not change in this process.

_Z^A\textrm{X}+_{-1}^0\textrm{e}\rightarrow _{Z-1}^A\textrm{Y}

The equation for the given reaction is,

_{19}^{40}\textrm{K}+_{-1}^0\textrm{e}\rightarrow _{18}^{40}\textrm{Ar}

Thus, the chemical symbol for the element that results from this process is, argon (Ar).

7 0
3 years ago
Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. x_{Ti} = 1.5

x_{O} = 3.5

%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

4 0
3 years ago
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