Mass of water 50.003 g 24 95C Temperature of water Specific heat capacity for water 4.184J/g C Mass of metal 3.546 Temperature o
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>
Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29)
Now, we use the definition of pH and clear [H₃O⁺] from there:
![pH=-log[H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M](https://tex.z-dn.net/?f=%20%5BH_3O%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D10%5E%7B-3%2C84%7D%3D0%2C00014%20M)
So, the [H₃O⁺] concentration is 0,00014 M
Have a nice day!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
Now, we use the definition of pH and clear [H₃O⁺] from there:
So, the [H₃O⁺] concentration is 0,00014 M
Have a nice day!
Answer:
1.06 mL of toluene will be needed.
Explanation:
Equi-molar mixture means equal moles of all the components.
as given the volume of ethyl acetate = 1mL
Density of ethyl acetate = 0.898 g/mL
The relation between density, mass and volume is :
mass=volumeXdensity
mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams
the moles are related to mass as:
For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol
moles of ethyl acetate will be:
So we need 0.01 moles of toluene also
For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene
mass required = 0.01 X 92=0.92grams
for 0.92 grams of toluene volume required will be: