Answer:
0.0559 mol
Explanation:
Step 1: Given and required data
- Volume of air (V): 1.35 L
- Pressure of air (P): 750 torr
- Ideal gas constant (R): 0.0821 atm.L/mol.K
Step 2: Convert "P" to atm
We will use the conversion factor 1 atm = 760 torr.
750 torr × 1 atm/760 torr = 0.987 atm
Step 3: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 17.0 °C + 273.15 = 290.2 K
Step 4: Calculate the number of moles of air
If we assume air behaves as an ideal gas, we can calculate the moles (n) of air using the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.987 atm × 1.35 L/(0.0821 atm.L/mol.K) × 290.2 K = 0.0559 mol
Answer:
The answer to your question is Rb₃N
Explanation:
Data
Lithium Nitride = Li₃N
Rubidium Nitride = ?
Process
1.- Both Lithium and Rubidium are placed in group lA of the periodic table, which means that their valence number if + 1.
2.- Nitrogen is placed in group VA, which means that its oxidation number is -3.
3.- Write the formula and interchange the oxidation numbers.
Rb⁺¹ N⁻³
Rb₃N₁
But number one is not necessary to write it.
Rb₃N
hi
1 mile is 1 609 meters. (rounded).
As there is 100 centimeters in a meter, then
1 mile is 160 900 centimeter.
Then 1.5 miles , is 1 mile + one half so :
160 900 + 160 900 /2 = 160 900 + 80 450
= 241 350
1.5 mile is 241 350 centimeters.
please note that the value of a mile was rounded.
Answer:
See Below.
Explanation:
This is a conversion problem:
Using the molar mass of Gold (given in a periodic table) which is 196.97g/mol
you have
you always arrange the equation in a way to cancel whatever you don't want (grams) and leave what you do want (moles). Here grams cancel (top and bottom), so you're left with:
moles of Gold
