Volume is the amount of space an object takes up
Answer:
1. 2 M
2. 2 M
Explanation:
1. Determination of the final concentration.
Initial Volume (V₁) = 2 L
Initial concentration (C₁) = 6 M
Final volume (V₂) = 6 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
6 × 2 = C₂ × 6
12 = C₂ × 6
Divide both side by 6
C₂ = 12 / 6
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
2. Determination of the final concentration.
Initial Volume (V₁) = 0.5 L
Initial concentration (C₁) = 12 M
Final volume (V₂) = 3 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
12 × 0.5 = C₂ × 3
6 = C₂ × 3
Divide both side by 3
C₂ = 6 / 3
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
The reaction 
→ 
is best classified as double displacement reaction.
Those reaction in which two compounds react by exchanges of ions to form two new compounds is called double displacement reaction. The easiest way to identify double displacement reactions is to check to see whether the cations exchanged anions with each other or not . Always balanced chemical equation is used to determine.
There are three types of double displacement reaction which is given as,
- Precipitation
- Neutralization
- Gas formation
The real world example of double displacement reaction is combining vinegar and baking soda to create homemade volcano.
learn more about double displacement reaction
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0.0760 m
do this by:
finding the moles of NaOH which will be <span>5.702 E -3 m
</span>
next find the moles of H3PO4 which will be <span>1.90 E -3 m</span><span>
calulcate </span>25 ml sample molarity = 0.07603 m, just put 0.0760<span>
</span>
Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
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