Answer:
Mass of water = 41.8 g
Explanation:
Given data:
Mass of water = ?
Change in temperature = 3.0 °C
Specific heat capacity = 4.184 j/g.°C
Heat absorbed = 525 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 3.0°C
Now we will put the values in formula.
525 J = m × 4.184 j/g.°C × 3.0°C
525 J = m × 12.552 j/g
m = 525 J/ 12.552 j/g
m = 41.8 g
Molar mass :
CO₂ = 44.0 g/mol
CH₄ = 16.0 g/mol
The balanced equation is wrong ,<span>this is the correct :</span>
<span>CH</span>₄<span> + 3 O</span>₂<span> = CO</span>₂<span> + 2 H</span>₂<span>O</span>₂<span>
</span>
16.0 g -------------> 44 g
184.5 g -----------> mass of methane
mass of methane = 184.5 x 44 / 16.0
mass of methane = 8118 / 16.0
mass of methane = 507.375
moles of Carbon dioxide ( CO₂)
507.375 / 44.0 => 11.531 moles of CO₂
hope this helps!
Answer:
There are many different types of preservatives like Benzoic acid, Calcium Sorbate, Erythorbic Acid, Potassium Nitrate and Sodium Benzoate. Some act like antioxidants used for slowing down spoilage like Ascorbyl Palmitate, Butylated Hydroxy anisole (BHA) and Butylated Hydroxytoluene (BHT
<u>Answer:</u> Copper (I) iodide will precipitate first.
<u>Explanation:</u>
We are given:
of CuCl = ![1.0\times 10^{-6}](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-6%7D)
of CuI = ![5.1\times 10^{-12}](https://tex.z-dn.net/?f=5.1%5Ctimes%2010%5E%7B-12%7D)
Concentration of ![Cl^-\text{ ion}=0.021M](https://tex.z-dn.net/?f=Cl%5E-%5Ctext%7B%20ion%7D%3D0.021M)
Concentration of ![I^-\text{ ion}=0.017M](https://tex.z-dn.net/?f=I%5E-%5Ctext%7B%20ion%7D%3D0.017M)
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
![K_{sp}=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
Putting values in above equation, we get:
![1.0\times 10^{-6}=[Cu^+]\times 0.021](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-6%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.021)
![[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-6%7D%7D%7B0.021%7D%3D4.76%5Ctimes%2010%5E%7B-5%7DM)
Concentration of copper (I) ion = ![4.76\times 10^{-5}M](https://tex.z-dn.net/?f=4.76%5Ctimes%2010%5E%7B-5%7DM)
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
Putting values in above equation, we get:
![5.1\times 10^{-12}=[Cu^+]\times 0.017](https://tex.z-dn.net/?f=5.1%5Ctimes%2010%5E%7B-12%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.017)
![[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B5.1%5Ctimes%2010%5E%7B-12%7D%7D%7B0.017%7D%3D3.00%5Ctimes%2010%5E%7B-10%7DM)
Concentration of copper (I) ion = ![3.00\times 10^{-10}M](https://tex.z-dn.net/?f=3.00%5Ctimes%2010%5E%7B-10%7DM)
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.
Its is true that light from an unshaded bulb radiates in all directions