1.more
2.longer
3.warmer
4.northern
5.less
6.shorter
7.colder
8.southern
Answer:
120 gram sample of a radioactive element, how many grams of that element will be left after 3 half-lives have passed? If you have a 300
Explanation:
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Answer:
The concentration of the analyte is determined by fitting the absorbance or transmittance obtained by spectrophotometric analysis of the unknown solution into the calibration curve.
Explanation:
In a calibration curve, the instrumental response (absorbance or transmittance), is plotted against the concentration of the analyte (the substance to be measured). The analyst is expected to prepare a series of standard solutions of the analyte within a range of solution concentrations close to the expected concentration of analyte in the unknown solution. The method of least squares may be used to determine the best fit of the line, thus, the concentration of the analyte. This method is only used for the determination of the concentration of coloured substances (spectrophotometry).
Atoms are the unit of the molecule of the compound. The 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water and 3 moles of carbon dioxide in the sample.
<h3>What are atoms?</h3>
Atoms are the smallest fundamental unit of the compounds that can be given by Avogadro's number.
For calculating the oxygen atoms in 5 mole water:
If 1 mole = 6.02 × 10²³
Then, 5 moles = 5 × 6.02 × 10²³
Hence, 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water.
Moles of carbon dioxide in the sample is calculated as:
If 1 mole of carbon dioxide = 6.02 × 10²³ molecules
Then moles in 1.8 x 10²⁴ molecules will be,
(1.8 x 10²⁴ molecules) ÷ (6.02 × 10²³ molecules) = 3 moles
Hence, 3 moles of carbon dioxide is present in the sample.
Learn more about Avogadro's number here:
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Answer:
Molarity = moles / volume
moles of HCl = molarity ×volume
= 0.0150 M × 0.08189L
= 0.0123mol
∴ the amount of CaCO3 neutralized by 0.0123 mole of HCl will be in 1:2 ratio.
moles of CaCO3 = 0.0123 mol of HCl× 1mol CaCO3/ 2mol HCl
= 0 .00614 mol
∴ the tablet has 0.00614 mol of CaCO3.
Explanation:
Step 1:
In the above-experiment problem, HCl acid solution is basically treated with CaCO3 base active in the tablet.
The mass of CaCO3 present in the tablet must be obtained in milligram (mg).
First, write down the balanced reaction between HCl and CaCO3
2Hcl aq + CaC03 s ⇒ CaCl2 aq + H2O l + C02 g
CaCO3 and HCl react in 1:2 ratio of moles respectively.
∴ for each mole of calcium carbonate there would be twice the moles of HCl required for neutralization.
Step 2
the volume is 81.89 mL (i.e 0.08189 L) of 0.150 M HCl neutralizes the whole tablet containing unknown amount of CaCO3.
So, the number of moles of HCl present in the taken volume is calculated
