Answer:
if a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one-half
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Answer:
Water - H2O
Ammonia - NH3
Sulfur dioxide - SO2
Hydrogen sulfide - H2S
Ethanol - C2H6O
Explanation:
Those are some atoms with polar covalent bonds. Hope this helps!!
Answer: 51 grams
Explanation:
Ammonia is a gas with a chemical formula of NH3.
Given that,
Amount of moles of NH3 (n) = ?
Volume of NH3 (v) = 200mL
since the standard unit of volume is liters, convert 200mL to liters
(If 1000mL = 1L
200mL = 200/1000 = 0.2L)
Concentration of NH3 (c) = 1.5M
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
Make n the subject formula
n = c x v
n = 1.50M x 0.2L
n = 3 moles
Now, calculate the mass of ammonia
Amount of moles of NH3 (n) = 3
Mass of NH3 in grams = ?
For molar mass of NH3, use the atomic masses:
N = 14g; H = 1g
NH3 = 14g + (1g x 3)
= 14g + 3g
= 17g/mol
Since, n = mass in grams / molar mass
3 moles = m / 17g/mol
m = 3 moles x 17g/mol
m = 51 grams
Thus, 51 grams of ammonia was dissolved in the solution.
1.75 moles ChCl3 x (6.02 x 10 ^-23) / 1 mole = 1.0535 x 10^-22 atoms.
hope this was helpful :)
