The mass of an ink pen with a triple beam balance if all the weights positioned to the right would be a maximum of 610 grams which sounds rather heavy for an ink pen but in the absence of necessary information it seems that that would have to be the answer.
Balancing of chemical equation is essential because of the law of conservation of mass, which states that the mass of a system can not be created or removed.
The second equation is balanced
This is because the number of elements of each atom in the product side equal the number of elements of each atom on the reactant side.
The first equation is not balanced
This is because there is 1 molecule of 
on reactant side as compared to 3 molecules of
To balance the equation we add a coefficient of 3 on sulphuric acid (
) and a coefficient of 3 on hydrogen (
)
Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
Best regards!
Atomic mass of helium is 4.002642g/mol
(542000g)/(4.002642g/mol)*6.02*10^23 = 8.15*10^28 atoms
Answer:
There are 3600 pages to 2 significant figures
Explanation:
Working only with the information given, we have that there are 23 books each with about 31 chapters.
This means that if 1 book has 31 chapter, the 23 science books will have
31 X 23 chapters =713 chapters in total.
If each chapter contains 5 pages, for the 713 chapters we will have a total of 713 X 5 pages = 3565 pages.
Rounding off the answer to 2 significant figures , we are looking to have only two non-zero figures in our answer. To do this, we count two numbers from the left. This will be numbers 3 and 5.
Next, we check the net number after 5 to see if it is greater than 5 if it is, we will have to add the number (1 ) to the 5. This shows that we are approximating upwards, to account for the fact that that number has passed the middle point.
After this, we change the remaining two numbers (6 and 5) to zeroes.
This will give us 3600 pages.
