Answer:
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions
Explanation:
We will first write the balanced equation for this scenario
3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2
3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2
The ratio here for both calcium chloride and magnesium nitrate is 
The number of moles of each compound is equal to
Using the mole ratio of 3:2, convert each to moles of sodium phosphate.
mole of CaCl2 is equal to 
Na3PO4
mole of CaCl2 is equal to 
Na3PO4
Converting moles of sodium phosphate to grams of sodium phosphate we get
g/mol
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions
I believe the best description comparing the structure of carbohydrates and nucleic acids would be A.
Answer:
about the same
Explanation:
Thermal energy is constant at a certain temperature in general. The energy is denoted by 
Where,
is the Boltzmann constant
T is the absolute temperature
Given that the balloon filled with nitrogen gas and the crystal of sugar are at room conditions means that they have same conditions and thus they will possess same energy irrespective of the states of the matter.
Thus,
The thermal energy of the sugar molecules is <u>about the same</u> as that of nitrogen molecules.
The sedimentary rock formed when water deposits tiny particles of clay in very thin flat layers is called limestone.
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
