Answer:
It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.
Explanation:
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
So, in this case:
- P= 1 atm
- V= 855 L
- n= ?
- R= 0.082
- T= 273.15 K
Replacing:
1 atm* 855 L= n* 0.082 * 273.15 K
Solving:
n= 38.17 moles
Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?
mass= 1,068.76 grams
<u><em>
It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>
Answer : The Lewis-dot structure of 
is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is,
As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in = 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
Hence, the Lewis-dot structure of is shown below.
Answer:
D
Explanation:
( I hope that this helps )
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation:
Answer: place a 2 in front of NaNo3 on left side of equation while leaving the other blanks empty or you can place a 1 in those blanks
Explanation:
Step 1 count and write down the amount of each given element for both sides
Step 2 begin placing numbers (coefficients) to each side to balance
