The more numbers after the decimal point there are, the more precise the instrument which recorded it is. For example, if one instrument during seismic activity records that the magnitude of the earthquake was 2.3, and another instrument recorded that it was 2.3645, the second instrument would have shown to be more precise.
Anna lives in a city that is part of the tropical climate types. It has a constantly warm weather, and thus higher humidity, and according to the annual rainfall, it is most probably a rainfall that appears seasonally, not throughout the whole year.
Tim, on the other hand, lives in a city that is part of the dry climate types. It is most probably a place that is deep into the mainland, like the cold deserts of Central Asia, where the temperatures in the summer are high, and in winter are very low. Because of the distance from the sea, the rainfall doesn't reach this places, so they are very dry, and only have symbolic amount of annual rainfall.
There is this thing called isotopes
it means that atoms of the same element can have a different number of neutrons.
if there is a change in the no. of neutrons, there will definitely be a change in the mass number.
so the answer is A) mass number
Answer: 1.14
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
To calculate pH of gastric juice:
molarity of 
= 0.072
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Thus the pH of the gastric juice is 1.14
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
