Answer:
approximately 15.1 grams.
Explanation:
The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.
change 25 grams of potassium chlorate to moles.
calculate the gram molecular mass of potassium chlorate.
Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.
1 K x 39 grams/mole
+1 Cl x 35.4 grams/ mole
+3 O x 16 grams/ mole
= 122.4 grams / mole Potassium Chlorate
25
122.4
= moles.
2.05 moles of Potassium Chlorate.
There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.
2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.
Find the gram molecular mass of Potassium Chloride.
1 K x 39 = 39
+1 Cl x 35.4 = 35.4
= 74.4 grams / mole.
2.05 moles x 74.4 grams/ mole = 15.2 grams
Explanation:
A. Hydrogen bonding is present in CS2 but not in CO2.
B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.
C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.
D. The dispersion forces are greater in CS2 than in CO2.
<u><em>PLS MARK BRAINLIEST :D</em></u>
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to 
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is 
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Answer:
a group of unicellular microorganisms
