Answer:
30.3 g
Explanation:
At STP, 1 mol of any gas will occupy 22.4 L.
With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:
- 32.0 L ÷ 22.4 L/mol = 1.43 mol
Then we <u>calculate how many moles would there be in 16.6 L</u>:
- 16.6 L ÷ 22.4 L/mol = 0.741 mol
The <u>difference in moles is</u>:
- 1.43 mol - 0.741 mol = 0.689 mol
Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 0.689 mol * 44 g/mol = 30.3 g
A) Sulfur dichloride (SCl₂) is polar molecule because is <span>bent with asymmetric charge distribution around the central atom (S).
b) </span>Sulfur tetrachloride (SCl₄) is polar because there is<span> lone electron pair around the sulfur.
c) </span>Bromine pentachloride (BrCl₅) is polar because dipole moment do not cancel.
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m
Answer:
electrophile(H⁺) is needed to react with alkene in the first step and nucleophile (OH⁻) is not available in the first step
Explanation:
Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.
Explanation:
Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.
Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .
Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.