Answer:
0.24%
Explanation:
A toothpaste contains 0.24% by mass sodium fluoride (NaF) used to prevent dental caries and 0.30% by mass triclosan, C12H7Cl3O2, a preservative and antigingivitis agent. One tube contains 119 g of toothpaste. (6.4, 6.5) Components in toothpaste include triclosan and NaF.a. How many moles of NaF are in the tube of toothpaste?b. How many fluoride ions (F?) are in the tube of toothpaste?c. How many grams of sodium ion (Na+) are in 1.50 g of toothpaste?d. How many molecules of triclosan are in the tube of toothpaste?Solution 124CQStep 1 of 4:Given:Mass of toothpaste = 119 g. Mass percent of sodium fluoride (NaF) = 0.24%Mass percent of triclosan (C12H7Cl3O2) = 0.30%a. Here, we are asked to find the number of moles of NaF in the tube of toothpaste.Mass percent of sodium fluoride (NaF) = 0.24%First, let’s find the mass of NaF in the tube: = 119 g = 0.2856 grams Molar mass of NaF = 41.98 g/mol This means, 1 mol NaF = 41.98 g. So, the mole-mass factor is: and Therefore, number of moles of NaF will be: = 0.2856 g NaF = 0.00680 mol NaFHence, the number of moles of NaF in 119g toothpaste is 0.00680 moles. ________________
Biosphere and Lithosphere
Answer:
Data obtained can't be accurate.
Explanation:
A beaker can't be used for measuring the volume of a liquid because they did not calculate volume of a liquid accurately and precisely. The marks present on the beaker are not accurate, it is just an estimate so by measuring the volume of any liquid using beaker give us a false data so that's why beaker are not used for the measurement of a volume.
Answer:
Ca
Explanation:
As a general trend on the periodic table, the electronegativities of elements decrease down the group and increase across the periods. Since calcium is in group two , it is expected to have the lowest amount of electronegativity due to its specific location on the periodic table.
To show that this prediction is true, I will write the elecronegativity values of these atoms. Calcium has a value 1.0, Silver 1.93 , Iron 1.83 and chromium 1.66
Answer:
a) K = 5.3175
b) ΔG = 3.2694
Explanation:
a) ΔG° = - RT Ln K
∴ T = 25°C ≅ 298 K
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔG° = - 4.140 KJ/mol
⇒ Ln K = - ( ΔG° ) / RT
⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))
⇒ Ln K = 1.671
⇒ K = 5.3175
b) A → B
∴ T = 37°C = 310 K
∴ [A] = 1.6 M
∴ [B] = 0.45 M
∴ K = [B] / [A]
⇒ K = (0.45 M)/(1.6 M)
⇒ K = 0.28125
⇒ Ln K = - 1.2685
∴ ΔG = - RT Ln K
⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )
⇒ ΔG = 3.2694