Answer:
Q.1
Given-
Volume of solution-1 L
Molarity of solution -6M
to find gms of AgNO3-?
Molarity = number of moles of solute/volume of solution in litre
number of moles of solute = 6×1= 6moles
one moles of AgNO3 weighs 169.87 g
so mass of 6 moles of AgNO3 = 169.87×6=1019.22
so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution
The fact that CO2 is released from oceans due to further rise in temperature is an example of a negative feedback.
A negative feedback is one in which the process that produces the feedback is interrupted. That is, the process is stopped as a result of the feedback received.
In this case, CO2 which leads to global warming dissolves in the ocean which serves a large sink for the gas. However, as the increase in ocean temperatures causes decrease in solubility of CO2, more CO2 is released leading to further temperature rise. This is an example of a negative feedback loop.
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43.8 has 3 significant figures and 1 decimal.
<h3 /><h3>What are significant figures?</h3>
The term significant figures refer to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation.
All zeros that occur between any two non-zero digits are significant. For example, 108.0097 contains seven significant digits. All zeros that are on the right of a decimal point and also to the left of a non-zero digit are never significant. For example, 0.00798 contained three significant digits.
Hence, 43.8 has 3 significant figures and 1 decimal.
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The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :
![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)
Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
Answer:
60-20=40km west because the direction Will favour whichever direction is bigger
