Answer:
A. 96.3 mg/dL
Absolute error: 5.7 mg/dL
Relative error: 5.6%
B. 97.2 mg/dL
Absolute error: 4.8 mg/dL
Relative error: 4.7%
C. 104.8 mg/dL
Absolute error: 2.8 mg/dL
Relative error: 2.7%
D. 111.5 mg/dL
Absolute error: 9.5 mg/dL
Relative error: 9.3%
E. 110.5 mg/dL
Absolute error: 8.5 mg/dL
Relative error: 8.3%
Explanation:
The formula for the absolute error is:
Absolute error = |Actual Value - Measured Value|
The formula for the relative error is:
Relative error = |Absolute error/Actual value|
In your exercise, we have that
Actual Value = 102.0 mg/dL
A. 96.3 mg/dL:


B. 97.2 mg/dL


C. 104.8 mg/dL


D. 111.5 mg/dL


E. 110.5 mg/dL


Answer:
Isotopes of an element have same number of protons but different number of neutrons. Which means isotopes of an element have same atomic number but different mass number.
The chemical property of an element is determined by the number of electrons. And as all the isotopes have same number of electrons, they have same chemical properties.
Thus as isotopes of an element have same atomic number , they have same number of electrons and protons. As they have different mass number, the number of neutrons will be different. Hydrogen has three isotopes ,
,
and
. Thus
has no neutron.
Answer:
pH = 3.65
Explanation:
given data
pKa of HNO2 = 3.40
nitrous acid (HNO2) = 0.110 M
NaNO2 = 0.200 M
to find out
What is the pH
solution
we get here ph for acidic buffer that is express as
pH = pKa + log(salt÷acid) ........................1
put here value and we get
pH = 3.40 + log(0.200÷0.110)
pH = 3.65
4 moles of hydrogen = 4 * 1.008 = 4.032 grams
1 mole of helium = 4.003 grams
Difference is 4.032 - 4.003
= 0.029 g
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2