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1 year ago

What would cause a nucleus to be unstable?

1 answer:

4 0

D

I don’t know how to explain it, but when there are 8 outermost outer nuclear electrons, it’s the most stable time, lose e- will make it unstable

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_________ is a systematic method of naming chemical compounds. This creates an unambiguous and consistent name for any chemical

Vlada [557]

Answer: IUPAC NOMENCLATURE

Explanation:

IUPAC stands for International Union of Pure and Applied Chemistry. They devised a systematic method for naming compounds in order to create a uniform global unambiguous system of nomenclature hence making it easier for researchers to share information more freely without the hindrance of reporting the same compound using different names in different parts of the world thus creating confusion in chemical literature.

6 0

3 years ago

7. A hypothesis needs to be written in a specific format.

Nastasia [14]

Answer:

a . knowledge and existing theories .

b. falsifiable scope

4 0

2 years ago

What is the bond angle in a tetrahedral molecule?

levacccp [35]

109.5

tetrahedral shape:
number of electron pair = 4,
number of bonded pair = 4,
number of lone pair = 0.

7 0

3 years ago

Read 2 more answers

PLEASE HELP! I really don’t understand! Thanks!

PIT_PIT [208]

http://century.rochester.k12.mn.us/cms/One.aspx?portalId=3086882&pageId=6133921

Hope this helps!

Please mark brainliest. :)

6 0

3 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

2 years ago