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
A physical property is what a substance is like; it's directly observable. On the other hand, a chemical property is how a substance behaves; its reactivity.
Examples of a physical property are: color, texture, boiling point, freezing point, and melting point.
Examples of a chemical property are: flammability, combustion, and formation of a precipitate.
Complete Question
The complete question is shown on the first question
Answer:
a) The duty of the heat exchanger is given as 6.8658 KJ /sec
b) The temperature of the water leaving the exchanger is TOUT = 29.84 ⁰C
c) The log mean difference is given as TZ = 47.317 ⁰ C
d) the UA value is UA = 145.10
Explanation:
The explanation is uploaded on the first and second ,third and fourth image
Answer:
half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive ...
Explanation:
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Answer:0.026ml
Explanation:
Details are found in the image attached. We must subtract the saturated vapour pressure of hydrogen gas at the given temperature from the total pressure of the hydrogen gas collected over water to obtain the actual pressure of hydrogen gas and substitute the value obtained into the general gas equation. The dry hydrogen gas has no saturated vapour pressure hence the value is substituted as given. All temperatures must be converted to Kelvin before substitution.
