Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.
Explanation:
i took the test and got it correct :) hope this helps
Answer:
Final pressure in (atm) (P1) = 6.642 atm
Explanation:
Given:
Initial volume of gas (V) = 12.5 L
Pressure (P) = 784 torr
Temperature (T) = 295 K
Final volume (V1) = 2.04 L
Final temperature (T1) = 310 K
Find:
Final pressure in (atm) (P1) = ?
Computation:
According to combine gas law method:
⇒ Final pressure (P1) = 5,048.18877 torr
⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760
⇒ Final pressure in (atm) (P1) = 6.642 atm
Always use least amount of sig figs possible. So this 9.7 would be (answer): 2 sig figs
91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Explanation:
2NaN3======2Na+3N2
This is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.
From the equation:
2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.
In the question moles of nitrogen produced is given as 2.104 moles
so,
From the stoichiometry,
3N2/2NaN3=2.104/x
= 3/2=2.104/x
3x= 2*2.104
= 1.4 moles
So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.
From the formula
no of moles=mass/atomic mass
mass=no of moles*atomic mass
1.4*65
= 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
