Answer:ohhhhhhh I’m confused on this
Explanation:ahhh
Answer:
C
Explanation:
Finding a new water source would be quite difficult
NaOH+ HCl -> NaCl+ H2O
Based on the above balanced equation, we know that 1 mol HCl= 1 mol NaOH (keep it in mind; it will be used later)
We know that 0.20M HCl solution (soln)= 0.20mol HCl/1L HCl soln (this one is based on the definition of molarity).
First, we have to find the mole of HCl:
5.0mL HCl solution* (1L soln/ 1,000mL soln)* (0.20mol HCl/1L HCl soln)= 1.0*10^(-3)mol HCl.
Now that we know the mole of HCl, let's find the molarity of NaOH:
1.0*10^(-3)mol HCl* (1mol NaOH/1 mol HCl)* (1/10.mL NaOH soln)* (1,000mL NaOH soln/1L NaOH soln)= 0.10mol/L NaOH soln = 0.10M NaOH soln
Hope this would help :))
Answer: 3.00 molecules
Explanation:
Given that:
Mass of CO2 = 132.03 grams
Number of molecules of CO2 = ?
Molar mass of CO2 = ?
For the molar mass of CO2: Atomic mass of Carbon = 12; Oxygen = 16
= 12 + (16 x 2)
= 12 + 32 = 44g/mol
Apply the formula:
number of molecules = (mass in grams/molar mass)
N = (132.03 grams/44g/mol)
N = 3.00 molecules
Thus, there are 3.00 molecules of CO2 in 132.03 grams of CO2.
Balanced equation:
4 FeS₂ + 11 O₂ --> 2 Fe₂O₃ + 8 SO₂
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Find the limiting reactant (the one that produces the least amount of product) :
(73.5 g FeS₂) * (1 mole FeS₂ / 119.975 g FeS₂) * (2 moles Fe₂O₃ / 4 moles FeS2) * (159.6882 g Fe₂O₃ / 1 mole Fe₂O₃)
= 48.9 g Fe₂O₃ …………. to 3 sig figs
(44 g O₂) * (1 mole O₂ / 31.9988 g O₂) * (2 moles Fe₂O₃ / 11 moles O₂) *
(159.6882 g Fe₂O₃ / 1 mole Fe₂O₃)
= 40 g Fe₂O₃ ………………. to 2 sig figs
O₂ is the limiting reactant and the theoretical yield is 44 g Fe₂O₃
