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kaheart [24]
3 years ago
11

Which particle has the same mass as a neutron?

Chemistry
2 answers:
Blizzard [7]3 years ago
6 0

Answer:

protons

Explanation:

approximately the same mass

Ilya [14]3 years ago
5 0

Answer:

proton

Explanation:

neutron: A subatomic particle forming part of the nucleus of an atom. It has no charge. It is equal in mass to a proton or it weighs 1 amu.

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Name the two types of sub-atomic particle in the nucleus of an atom.
Talja [164]
The name of the sub-atomic particles in the nucleus of atom are protons and neutrons


Hope it helps you!!!!
5 0
3 years ago
Which of the following solutions is a good buffer system?A solution that is 0.10 M HCN and 0.10 M NaClA solution that is 0.10 M
patriot [66]

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

Explanation:

  • A good buffer system contains a weak acid and its salt or a weak base and its salt.
  • In this case; A solution that is 0.10 M HCN and 0.10 M LiCN, would make a good buffer system.
  • HCN is a weak acid, while LiCN is a salt of the weak acid, that is, CN- conjugate of the acid.
8 0
3 years ago
URGENT <br> Best answer gets Brainliest!<br> Mg+ 2HCI ——&gt; MgCI2+ H2
avanturin [10]

Answer:

Reaction type: Single displacement

Reactant: Magnesium

Product: Dihydrogen - H2

6 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
What is the blocks average speed nearest hundredth of a m/s
bezimeni [28]

The question is incomplete, the complete question is;

A block is pulled 0.90 m to the right in 2.5 s. What is the blocks average speed to the nearest hundredth of a m/s

Answer:

0.36 m/s to the hundredth place

Explanation:

Now let us remember the definition of speed. Speed is defined in physics as distance/time.

Here we have the distance as 0.9 m

We have the time as 2.5 s

Hence the average speed is obtained from;

Speed = 0.9/2.5 = 0.36 m/s to the hundredth place

6 0
2 years ago
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