Uno:
El dióxido de Carson (CO2)
A.how many moles of aspirin are in one tablet
0.325 g ( 1 mol / 180.2 g ) = 1.80 x 10^-3 mol
B. How many moles of oxygen are in one tablet of aspirin
1.80 x 10^-3 mol C9H8O4 ( 4 mol O / 1 mol <span>C9H8O4) = 7.21 x 10^-3 mol O
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C. How many atoms of carbon are in one tablet of aspirin
1.80 x 10^-3 mol C9H8O4 ( 9 mol C / 1 mol C9H8O4 ) ( 6.022 x 10^23 atoms C / 1 mol C ) = 9.76 x 10^21 atoms C
D. How many grams of oxygen in one tablet of aspirin
7.21 x 10^-3 mol O ( 16 g / mol ) = 0.12 g O
Answer:
True
Explanation:
There are two types of Van der waals forces which are intermolecular bonds. These bonds are London dispersion forces and Dipole-dipole attractions.
The london dispersion forces are weak attractions found between non-polar molecues and noble gas.
The attractions here is as a result of the fact that non-polar molecules or atoms sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. This leads to the formation of a temporary dipole or instantaneous dipole. The dipole can induce the neighbouring molecules to be distorted and they form dipoles as well.
Answer:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u>
The final concentration of sodium anion = 0.1035 M</u>
Answer:
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