Answer: Its average atomic mass is 114.9 amu
Explanation:
Mass of isotope 1 = 113 amu
% abundance of isotope 1 = 5% = 
Mass of isotope 2 = 115 amu
% abundance of isotope 2 = 95% = 
Formula used for average atomic mass of an element :
![A=\sum[(113\times 0.05)+(115\times 0.95)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%28113%5Ctimes%200.05%29%2B%28115%5Ctimes%200.95%29%5D)
Thus its average atomic mass is 114.9 amu
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
if the scientist finds anything that does not match, they have to carry out further tests
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Answer:the minor component in a solution, dissolved in the solvent.
Explanation:
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