What would happen if the number of protons were to change in an atom?

1 answer:

5 0

Answer: you will change the atom from one element to a different element. Sometimes, when you add a proton to an element, the element will become radioactive. If you change the number of neutrons in an atom, you get an isotope of the same element.

Explanation:

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Some species are no longer living but we know they existed one time. we know this because....

Dahasolnce [82]

We know this because of the discovery of fossils and bones. We can find out roughly when they existed because of carbon dating.

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3 years ago

Isobutylene is a hydrocarbon used in the manufacture of synthetic rubber. when 0.847 g of isobutylene was subjected to combustio

Firlakuza [10]

Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.

Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.

Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.

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3 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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3 years ago

Which statement applies to electronegativity?

densk [106]

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Electronegativity is defined as the tendency of an atom to attract the shared pair of electrons towards itself whenever a bond is formed.

This property increases as we move from left to right across a period because the number of charge on the nucleus gets increased and electrons are attracted more towards the nucleus.

This property decreases as we move from top to bottom in a group because the electrons get add up in the new shells which make them further away from the nucleus.

Thus, the correct answer is Option A.

7 0

2 years ago

Chemistry Lab Question: Filtration

tatyana61 [14]

For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.

For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

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3 years ago