Explain why we see the pattern between predator and prey
1 answer:
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Answer:
Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.
The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.
Multiply on the right side of the equation. The units of degrees Celsius cancel.
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g
The units of Joules cancel.
The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.
The mass of the sample of metal is approximately <u>333 grams.</u>
The question is incomplete, the complete question is attached below.
Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
Explanation : Given,
Mass of NaOH = 60 g
Volume of stock solution = 300 mL
Molar mass of NaOH = 40 g/mol
First we have to calculate the molarity of stock solution.
Now put all the given values in this formula, we get:
Now we have to determine the volume of stock solution and distilled water mixed.
Formula used :
where,
are the molarity and volume of stock solution.
are the molarity and volume of diluted solution.
From data (A) :
Putting values in above equation, we get:
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (B) :
Putting values in above equation, we get:
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (C) :
Putting values in above equation, we get:
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From data (D) :
Putting values in above equation, we get:
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.
Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
