Answer:
The 2 would be placed in front of the reactant Na and in front of the product NaCl
2Na + Cl2 = 2NaCl
Explanation:
This is because the ratio of elements needs to be balanced on both sides.
On the reactants side, there are 2 Na molecules and 2 Cl molecules
On the products side, there are 2 Na molecules and 2 Cl molecules
So, now the equation is balanced
If you mean hydrate as in <em>MgSO4 · 7H2O, </em>then simply find the molar mass of each element you see.
For the example above, that means you would add the molar mass (found on the periodic table) of Mg, then S, then 4(O), 14(H), and 7(O).
The results would be your molar mass for the hydrate.
I hope this is what you meant by your question!
<u>Answer:</u> The final volume of the gas comes out to be 4 L.
<u>Explanation:</u>
To calculate the volume with changing pressure, we use the equation given by Boyle's law.
This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
Mathematically,
(At constant temperature and number of moles)
The equation given by this law is:
where,
are initial pressure and volume.
are final pressure and volume.
We are given:
Putting values in above equation, we get:
Hence, the final volume of the gas will be 4 L.
<span>
<span>
</span></span>Volume = 4/3 * PI * r^3
1.59 x 10^24 copper atoms = 2.64 moles of copper
Atomic Mass of copper = 63.55
2.64 * 63.55 = 167.77 grams of copper
Volume of Copper = Mass / Density
Volume of Copper = 167.77 grams / 8.96
Volume of Copper = <span>
<span>
18.72</span></span> cubic centimeters
r^3 = Volume / (4/3 * PI)
r^3 = 18.72 / 4.188
r^3 =
<span>
<span>
<span>
4.47
radius = </span></span></span><span><span><span>1.647</span> centimeters
</span></span>
<span>Surface tension is most prevalent in earth's most important compound, water. "Normal" water at ambient temperature has a high surface tension, but the addition of different chemicals can alter this. The relationship between the molecules affects the fluidity of the liquid and it's ability to support pressure</span>
