first we have to find the empirical formula of the compound
empirical formula is the simplest ratio of whole numbers of components making up a compound
for 100 g of the compound
N O
mass 46.7 g 53.3 g
number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol
moles = 3.34 mol = 3.33 mol
divide by the least number of moles
3.34/3.33 = 1.00 3.33/ 3,33 = 1.00
therefore number of atoms are
N - 1
O - 1
empirical formula is - NO
mass of empirical unit - 14 g/mol + 16 g/mol = 30 g
molecular formula is actual composition of elements in the compound
molecular mass - 60.01 g/mol
number of empirical units = molecular mass / empirical unit mass
= 60.01 g/mol / 30 g = 2
there are 2 empirical units
2(NO)
molecular formula = N₂O₂
Aqueous solutions of barium nitrate and potassium phosphate are mixed.
What is the precipitate and how many molecules are formed?
Barium nitrate has a chemical symbol of Ba(NO3)2 and potassium phosphate
has a chemical symbol K2PO4. The reaction between these two is a double
replacement reaction yielding barium phosphate and potassium nitrate.
The chemical equation representing the reaction is,
Ba(NO3)2 + K2PO4 à KNO3 +
BaPO4
Climate is considered an abiotic limiting factor because it is non living . hope this helped :))
Answer:
Option =C each mouse in the study has brown ears.
Explanation:
Qualitative:
Qualitative properties are those that can be only observed but not measured in numerical values. These are observed through senses: touch, sight, smell, taste and hear.
For example:
Color, odor, brittleness, taste etc.
Quantitative:
Quantitative properties can be measured in numerical values.
For example:
Melting point, boiling point, conductivity, viscosity, density, hardness and solubility.
In short we can say that qualitative is a measure of quality while the quantitative is a measure of quantity.
Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample = 
= 
(where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL
= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL
