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3 years ago

If the plum pudding model of the atom was correct, what should the results of Rutherford’s experiment be?

1 answer:

6 0

Answer:most of the positively charge particles should be bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil

Explanation:

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Based on the observations of how the different forms of salt start mixing with the

zimovet [89]

I would expect fine salt to fully dissolve by the end of 45 minutes. Since the other types of salt are not fine and tiny, they would take longer to dissolve

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3 years ago

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HELP ASAP!!!!!!!!!!

weqwewe [10]

Answer:

D) food that a chef seal in a plastic bag and cook under a controlled temperature condition.

Explanation:

Sous vide, also known as low temperature long time cooking, is a method of cooking in which food is placed in a plastic pouch or a glass jar and cooked in a water bath for longer than usual cooking times at a precisely regulated temperature.

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3 years ago

Carbohydrates are formed by plants converting water a d carbon dioxide into glucose and oxygen, in the photocatalyzed process is

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Photosynthesis maybe.

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2 years ago

An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having

dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass 2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007 (1-x) = 1.008

2.014x + 1.007 - 1.007 x = 1.008

2.014x - 1.007x = 1.008 - 1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

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3 years ago

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

Leto [7]

Answer: The nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

$_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}$

For the given equation :

$^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}$

As the atomic number and mass number must be equal on both sides of the nuclear equation:

$^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}$

Thus the nuclide symbol of X is $^{231}_{90}\textrm{Th}$

5 0

3 years ago