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3 years ago

Developing Scientific knowledge

Chemistry

1 answer:

Talja [164]3 years ago

7 0

Answer:

Scientists develop knowledge by making observation about the natural world that may lead to a scientific question.

A scientific question may lead to a(n) hypothesis which can be tested.

The results of experimentation can lead to changes in scientific knowledge.

Explanation:

In the process of developing scientific knowledge, scientists carry out series of observations on what happens in the natural world. This helps them to develop scientific questions. These scientific questions drawn can then lead to hypotheses which are testable.

The testing process leads to experimentation. It is in this process that much scientific questions receive answers. Testing actually lies at the core of scientific inquiry for all scientists. Whatever hypothesis made must always be tested.

The result of the testing or experimentation brings changes in scientific knowledge.

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Solid potassium chlorate (kclo3) decomposes into potassium chloride and oxygen gas when heated. how many moles of oxygen form wh

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4 years ago

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

Answer: The average atomic mass of element Zinc is 65.40 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

For $_{30}^{67}\textrm{Zn}$ isotope:

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

For $_{30}^{68}\textrm{Zn}$ isotope:

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

For $_{30}^{70}\textrm{Zn}$ isotope:

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

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4 years ago

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4vir4ik [10]

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3 years ago

How many grams of water when supplied with 348j of heat will gain a temperature of 5.2°c?

Zielflug [23.3K]

We can use the heat equation,

Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

In this problem there is no any data about initial temperature of the water. So, we can assume that given temperature of 5.2 °C as the temperature difference.

Q = 348 J

m = ?

c = 4.186 J g⁻¹ °C⁻¹

ΔT = 5.2 °C

By applying the formula,
348 J = m x 4.186 J g⁻¹ °C⁻¹ x 5.2 °C
m = 15.99 g

Hence, the grams of water is 15.99.

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3 years ago

Proteins...

Bess [88]

All of the above (#4)

5 0

3 years ago