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densk [106]
3 years ago
7

Which two particles each have a mass approximately equal to one atomic mass unit?

Chemistry
2 answers:
natali 33 [55]3 years ago
3 0

Answer is: (4) proton and neutron.

One unified atomic mass unit is approximately the mass of one nucleon (proton or neutron).

The unified atomic mass unit (amu) is a standard unit of atom mass.  

One unified atomic mass unit is approximately the mass of one nucleon (1.66·10⁻²⁷ kg).

For example, one carbon atom has mass of 12 amu (unified atomic mass).

The electron has a mass that is approximately 1/1836 that of the proton.

GarryVolchara [31]3 years ago
3 0

Answer : The correct option is, (4) proton and neutron

Explanation :

There are three subatomic particles which are protons, neutrons and electrons.

The protons and the neutrons are present in the nucleus. Protons are positively charged and neutrons has no charge.

The electrons are present around the nucleus or outside the nucleus and electrons are negatively charged.

The mass of proton and neutron is equal to one atomic mass unit.

The mass of electron is zero.

The positron is also a subatomic particle with the same mass of an electron. But the positron are positively charged. That means the mass of positron is also zero.

Hence, the proton and neutron two particles each have a mass approximately equal to one atomic mass unit.

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During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

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2 years ago
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Answer:

B

Explanation:

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Answer:

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