Answer:
There are three atoms in the equation
I think the most appropriate answer is: the solvent being used in the experiment
<span>To correct for any light absorption not originating from the solute you will need to calibrate the tools with a solution that most similar to the sample.
Blank covete or standard solution can be used, but it was not ideal. By using the solvent as calibration, you can remove the reading from the solvent so your result only comes from the sample.
</span>
Answer:
B.) 129.9 grams
Explanation:
To find the mass, you need to use the following equation:
Q = mcΔT
In this equation,
-----> Q = energy (J)
-----> m = mass (g)
-----> c = specific heat (J/g°C)
-----> ΔT = change in temperature (°C)
The specific heat of copper is 0.385 J/g°C. Knowing this, you can plug the given values into the equation and simplify to isolate "m".
Q = mcΔT <----- Equation
5000 J = m(0.385 J/g°C)(200 °C - 100 °C) <----- Insert values
5000 J = m(0.385 J/g°C)(100) <----- Subtract
5000 J = m(38.5) <----- Multiply 0.385 and 100
129.9 = m <----- Divide both sides by 38.5
Answer: The rate of appearance of
is 
Explanation:
Rate of a reaction is defined as the rate of change of concentration per unit time.
Thus for reaction:

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1d%5BI%5E-%5D%7D%7B5dt%7D%3D%2B%5Cfrac%7Bd%5BI_2%5D%7D%7B3dt%7D)
Given:
= 
![+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls](https://tex.z-dn.net/?f=%2B%5Cfrac%7Bd%5BI_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B3d%5BI%5E-%5D%7D%7B5dt%7D%3D-%5Cfrac%7B3%7D%7B5%7D%5Ctimes%202.4%5Ctimes%2010%5E%7B-3%7Dmol%2FLs%3D1.44%5Ctimes%2010%5E%7B-3%7Dmol%2FLs)
The rate of appearance of
is 
Answer:
The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ
Explanation:
<u>Step 1: </u>The balanced equation
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH = -802 kJ
<u>Step 2:</u> Given data
We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.
The enthalpy change of combustion, given here as Δ
H
, tells us how much heat is either absorbed or released by the combustion of <u>one mole</u> of a substance.
In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number), 802.3 kJ of heat.
<u>Step 3: </u>calculate the enthalpy change for 3 moles
The -802 kj is the enthalpy change for 1 mole
The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ
The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ