Answer:
67.91 g of CuCl2; 32.09 g of Cu.
Explanation:
The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.
Answer:
0.464 L
Explanation:
Molarity (M) = number moles (n) ÷ volume (V)
According to the information given in this question:
number of moles (n) = 4.36 moles
Molarity = 9.4M
Volume = ?
Using M = n/V
9.4 = 4.36/V
9.4V = 4.36
V = 4.36/9.4
V = 0.464 L
Hence, 0.464L of water are needed the volume of water.
Explanation:
The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.
Total pressure = 2atm
Mole Fraction = number of moles / total number of moles
Neon
Mole Fraction = 4.46 / 7.35 = 0.607
Partial Pressure = 0.607 * 2 = 1.214 atm
Argon
Mole Fraction = 0.74 / 7.35 = 0.101
Partial Pressure = 0.101 * 2 = 0.202 atm
Xenon
Mole Fraction = 2.15 / 7.35 = 0.293
Partial Pressure = 0.293 * 2 = 0.586 atm
<span>A flashlight is an electric-powered light source; the light source is a light bulb or an LED. The electrical energy is converted into visible light. Flashlights can be hand-held or mounted to a platform. Light from a lighting, on the other hand, is formed by exciting electrons to a higher state. </span>
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
