aleksA sailor on a trans-Pacific solo voyage notices one day that if he puts of fresh water into a plastic cup weighing , the cu
p floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right). Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to significant digits. You'll need to know that the density of fresh water at the temperature of the sea around the sailor is . You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
1 answer:
Answer:
40 g
See explaination
Explanation:
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
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Answer: a. Synthesis
Explanation:
a. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.
Example:
Thus magnesium in its elemental form is combining with oxygen to form magnesium oxide.
b. Double displacement reaction is one in which exchange of ions take place.
Example:
c. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.
Example:
<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
<u>For NaOH:</u>
Initial molarity of NaOH solution = 0.195 M
Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
<u>For sulfuric acid:</u>
Initial molarity of sulfuric acid solution = 0.248 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
The chemical equation for the reaction of NaOH and sulfuric acid follows:
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, moles of KOH will react with =
of sulfuric acid
As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of sulfuric acid =
By Stoichiometry of the reaction:
2 moles of KOH produces 1 mole of sodium sulfate
So, moles of KOH will produce =
of sodium sulfate
- <u>For sodium sulfate:</u>
Moles of sodium sulfate =
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
- <u>For sulfuric acid:</u>
Moles of excess sulfuric acid =
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
- <u>For NaOH:</u>
Moles of NaOH remained = 0 moles
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.