Hey there!:
<span>Reaction stoichiometry :
</span>
Number of moles NaHCO3 :
Molar mass NaHCO3 = <span>84.007 g/mol
</span>
n = m / mm
n = 2.50 / 84.007
n = 0.0297moles of NaHCO3
2 NaHCO3 + H2SO4 = Na2SO4 + 2 CO2 + 2 H2O
2 moles NaHCO3 ----------------- 1 mole H2SO4
0.0297 moles NaHCO3 ----------- moles H2SO4
moles H2SO4 = 0.0297 * 1 / 2
moles H2SO4 = 0.0297 / 2
= 0.01485 moles of H2SO4
Therefore:
Molarity ( H2SO4 ) = moles H2SO4 / volume
0.600 M = 0.01485 / V
V = 0.01485 / 0.600
V = 0.02475 L of H2SO4
hope this helps!
Answer:A
Explanation: because i saw the sheet
Answer:
just be yourself , if they are a good person that you want to be your friend they would support you and care about what you have to say.
Explanation:
Answer:
1.8 g
Explanation:
Step 1: Write the balanced equation
CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.
The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.
Thus, the limiting reactant is CH₃CH₃
Step 3: Calculate the mass of CO₂ produced
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.
0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g
<span>34 g/mol if using the post 1981 definition of STP
32 g/mol if using the pre 1982 definition of STP
One minor complication is that the definition of STP changed in 1982 and some text books still use the old definition. So for this problem, I'll be giving two answers, but showing the work for only one of the results.
Pre 1982, STP was defined as 273.15 K at exactly 1 atmosphere (101325. Pascals)
Post 1981, STP was defined as 273.15 K at exactly 100 kPa (100000 Pascals)
The result of this difference is that the volume of 1 mole of a gas at STP differs.
Pre 1982, 1 mole of gas at STP has a volume of 22.414 liters.
Post 1981, 1 mole of gas at STP has a volume of 22.71098 liters.
First, determine the number of moles of gas molecules we have.
19.44 l / 22.71098 l/mol = 0.85597363 mol
Now determine how many moles of Ar, and Ne we have. Look up the atomic weights.
Atomic weight Argon = 39.948
Atomic weight Neon = 20.1797
Moles Argon = 8.2 / 39.948 = 0.205266847 mol
Moles Neon = 8.2 / 20.1797 = 0.406348955 mol
Calculate the number of moles of unknown gas by subtracting the number of moles of argon and neon we have from the number of moles of gas in total. So
0.85597363 mol - 0.205266847 mol - 0.406348955 mol = 0.244357829 mol
Now that we know how many moles of the unknown gas we have, we can simply divide the mass by the number of moles, so
8.2 g / 0.244357829 mol = 33.55734514 g/mol
Rounding to 2 significant figures gives a molar mass of 34 g/mol and since there isn't any element with a atomic weight of 17, I suspect that the textbook you're using is using the pre-1982 definition of STP. So repeating the calculations above with a volume of 22.414 liters per mole (pre-1982 definition of STP), the newly calculated molar mass becomes 32 g/mol which is a nice match for the unknown gas being oxygen.
The differences using the older standard are:
19.44 l / 22.414 l/mol = 0.867315071 mol ; Number of moles in the volume of 19.44 l
0.867315071 mol - 0.205266847 mol - 0.406348955 mol = 0.255699269 mol ; moles of unknown gas.
8.2 g / 0.255699269 mol = 32.06892229 g/mol ; Molar mass of unknown gas.</span>