Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, 
.
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
The given reaction equation is as follows.
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, .
Answer:
The empirical formula is Ag2O.
The empirical formula is Ag2O.Explanation:
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.
do the steps ...
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
Answer:
63. 55 amu
Explanation:
Copper is known to exist in two different isotopes which are Cu-63 and Cu-65.
Cu-63 has an atomic mass of 62.93 amu and it has an abundance of 69.15%.
Similarly,
Cu-65 has an atomic mass of 64.93 amu and it has an abundance of 30.85%
Therefore, using the weighted average mass method, the atomic mass of copper is:
Atomic mass of copper = (0.6915*62.93) amu + (0.3085*64.93) amu = 43.52 amu + 20.03 amu = 63.55 amu
Thus, the atomic mass of copper (express in two decimal places) is 63.55 amu
It is C 2 1 2 1
You have to 1st balance the nitrates then balance the silvers to get the coefficients
Answer:
Take a look at the attachment below
Explanation:
Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.
Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.
<em>Hope that helps!</em>
