Easy stoichiometry conversion :)
So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.
So, our first step would look like this:
10.0
------
1
Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.
So, our 2nd step would look like this:
1 mole CO2
-----------------
84.007g NaHCO3
When we put it together: our complete stoichiometry problem would look like this:
10.0g NaHCO3 1mol CO2
---------------------- x -------------------------
1 84.007g NaHCO3
Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)
And then....
Divide the top answer by the bottom answer.
10.0/84.007 is 0.119
So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.
Hope I could help!
1.5 x 10⁶mm³
Explanation:
Here we are to convert m³ to mm³ or cm³ to see the correct option.
100cm = 1 m
1000mm = 1m
Converting to cm³ from m³;
0.0015m³ = 1.5 x 10⁻³m³
Now:
1.5 x 10⁻³ x m x m x m x
x
x 
= 1.5 x 10³cm³
Also;
1.5 x 10⁻³ x m x m x m x
x
x 
= 1.5 x 10⁶mm³
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The following of what exactly?
Answer :
V - mass of the water in the calorimeter
V - mass of the metal
V - change in temperature of the water
V - change in temperature of the metal
C - volume of water in calorimeter
C - calorimeter pressure
C - specific heat of water
Explanation :
Variables : It is a factor that changes during the experiment or calculation.
Constant : It is a factor that does not change during the experiment or calculation.
In a calorimeter, the heat absorbed is equal to the heat released.

As we know that,

![m_1\times c_1\times \Delta T_1=-[m_2\times c_2\times \Delta T_2]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%5CDelta%20T_1%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%5CDelta%20T_2%5D)
where,
= mass of water in calorimeter
= mass of metal
= change in temperature of the water
= change in temperature of the metal
= specific heat of water
= specific heat of metal
From this, we conclude that the value of specific heat of water is constant while the other are variables.
The volume of water in calorimeter, calorimeter pressure is also constant.