Answer: The number of grams of
in 1620 mL is 1.44 g
Explanation:
According to ideal gas equation:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)
n = number of moles = ?
R = gas constant =![0.0821Latm/Kmol](https://tex.z-dn.net/?f=0.0821Latm%2FKmol)
T =temperature =![273K](https://tex.z-dn.net/?f=273K)
![n=\frac{PV}{RT}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D)
![n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1atm%5Ctimes%2016.2L%7D%7B0.0821Latm%2FK%20mol%5Ctimes%20273K%7D%3D0.72moles)
Mass of hydrogen =![moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g](https://tex.z-dn.net/?f=moles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.72mol%5Ctimes%202g%2Fmol%3D1.44g)
The number of grams of
in 1620 mL is 1.44 g
Answer:
Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.
Explanation:
According to Bronsted-Lowery concept of acid and base, Acid is that substance which donated H⁺, while, Base is that substance which accepts H⁺.
Hydroiodic Acid being a strong Acid with Ka value of 3.2 × 10⁹ is when dissolved in water it donates H⁺ to water and water acts as a Base. i.e.
HI + H₂O → H₃O⁺ + I⁻
H₃O⁺ formed is the conjugate Acid of H₂O, while I⁻ is the Conjugate Base of HI.
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm