<h2>Answer:</h2>
The density of mercury molecule is higher than water.
<h3>Explanation:</h3>
Density is defined as mass per unit volume.In other words, density is the amount of matter within a given amount of space. water has the density of 1.0 gram per milliliter whereas the mercury has a density of 13.6 grams per centimeter squared.
One reason for the differences in density between mercury and water is that the atomic mass of mercury is 200.59 grams per mole. The atomic mass of water is 18.0 grams per mole. This is because mercury has a larger nucleus than hydrogen or water.
Additionally, there are strong inter-molecular forces (hydrogen bonds) between water molecules. hydrogen molecules do not stack upon one another as nicely as mercury atoms. Thus, there is additional empty spaces between the water molecules leading to its lower mass per volume(density)
Answer:
Explanation:
a. CuO+ 2HCl⇒CuCl2+ H2O
b.
=
= 0,05 (mol)
⇒
=
=0,05 mol
⇒
= 0,05×135=6,75 (g)
c.
=2×
=0,1 (mol)
⇒
= 0,1×36,5= 3,65 (g)
⇒
=
×100=36,5 (g)
⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=
×100=
×100≈ 16,67%
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
Explanation:
The number of moles of solute present in liter of solution is defined as molarity.
Mathematically, Molarity = 
Also, when number of moles are equal in a solution then the formula will be as follows.

It is given that
is 8.00 M,
is 7.00 mL, and
is 0.80 M.
Hence, calculate the value of
using above formula as follows.



= 70 ml
Thus, we can conclude that the volume after dilution is 70 ml.
The answer is "Incidents"