A. A pacemaker
Hope this helps!
Answer:
A) yes
B) The average mass of 12 pennies should be expressed as
( 3.131 + 3.129 + -------- X12 ) g / 12
where X12 = is the mass of the 12th penny
Explanation:
A) I think the the Bureau of Mint changed the way it made pennies because from the experiment and observation carried out on 15 different pennies it can be seen that they had different weights and this difference is associated with the period/time of production of the different pennies,
B ) The average mass of any object should be expressed in the si unit of mass and not having ($) sign attached to the value instead it should be expressed in grams or kilograms (s.i unit of mass )
The average mass of 12 pennies should be expressed as
( 3.131 + 3.129 + -------- X12 ) g / 12
where X12 = is the mass of the 12th penny
Answer:
K loses one electron to CI
Explanation:
The lewis electron dot notation shows only the chemical symbol of the element surrounded by dots to represent the valence electrons.
We have atom of K with one valence electrons
Cl with 7 valence electrons
For an electrostatic attraction to occur, both particles must be charged. To do this, one of the species must lose an electron, and the other gains it.
This will make both species attain a stable octet;
Hence, K will lose 1 electron and Cl will gain the electrons.
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
