There are 0.462 moles of atoms in 6.00 g of carbon-13.
Molar mass = mass of a substance/ Number of moles
g/mol = g /mole
M = m/n; where M is the molar mass, m is the mass and n is the number of moles
We have; n = m/M
In our case;
Mass = 6 g of Carbon-13
Molar mass = 13.0 g/mol
Since; Number of moles = Mass/ molar mass
Thus;
Moles = (6.0 g)/ (13.0 g/mol)
= 0.462 moles
Keywords: Moles, Molecular mass, relative atomic mass
Level: High school
Subject: Chemistry
Topic: Moles
Sub-topic: Moles, molecular mass and mass of a pure substance
Answer: 0.462 moles
Explanation: 13C indicates an isotope of carbon and its mass number is 13. It means the mass of 1 mol of 13C is 13 gram.
The question asks to calculate the number of atoms present in 6.00 grams of 13C.
To calculate the number of moles we divide the given grams by the mass of 1 mol of the element. The set could be shown easily using dimensional analysis as:
= 0.462 moles
So, there will be 0.462 moles of atoms in 6.00 grams of 13C.
(2) radium-226 (4) thorium-232
Answer: Option (1) is the correct answer.
Explanation:
A positron is represent by the symbol .
For example,
The decay reaction is as follows.
The reaction will be as follows.
Therefore, we can conclude that out of the given options positrons are spontaneously emitted from the nuclei of potassium-37.
b. 3KCl + HNO3 - > 3HCl + KNO3
c. 2Be + O2 - > 2BeO
d. Mg + O2 - > MgO2
this question two answers which is C and D they are balanced to the right tab.
B. Salinity
C. Soil depth
D. Sunlight
E. Temperature
a. CH4 and C2H4
b. PbCl2 and PbCl4
c. N2O5 and NO2
d. C2H6 and C4H12
Answer: The correct option is d.
Explanation: Empirical formula is a chemical formula which has the simplest ratio of elements present in a chemical compound.
From the given following pairs, the pair which shares the same empirical formula is
Empirical formula of is (by dividing the coefficients of by 3)
Empirical formula of is (by dividing the coefficients of by 3)
Both the chemical formulas have same empirical formula. Hence, the correct option is d.