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2 years ago

Neutron notes

Chemistry

1 answer:

Montano1993 [528]2 years ago

7 0

Answer:

A neutron is a subatomic particle with a mass comparable to a proton and a neutral electrical charge. A neutron is a subatomic particle with a mass of 1 and charge of 0. ... While the number of protons in a atom determines its element, the number of neutrons determines its isotope

Explanation:

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What happens to the number of neutrons as you move across the table? <br><br> left to right

luda_lava [24]

Answer: The number of neutrons will increase as we move from left to right in a periodic table.

Explanation:

Atomic number is equal to the number of protons.

Mass number is the sum of number of neutrons and number of protons.

As we move from left to right, both the atomic number and mass number increases.

For example: As we move from Lithium to berrylium to boron to carbon to nitrogen to oxygen to fluorine to neon , the number of neutrons increase from 4 to 5 to 6 to 6 to 7 to 8 to 10 to 10.

Thus the number of neutrons will also increase as we move from left to right in a periodic table.

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What is the formula of the compounds Cl- and Na+

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Na and Cl together form the compound
NaCl2

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3 years ago

What property of water enables ice wedging to break rock?

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Answer:

Explanation:

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2 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.

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2 years ago