its 40
I just took the test right now to get this answer
V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L
M ( HCl ) = ?
V ( NaOH ) = 25.00 / 1000 => 0.025 L
M ( NaOH) = 0.2000 M
number of moles NaOH :
n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH
Mole ratio:
HCl + NaOH = NaCl + H2O
1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH
moles HCl = 0.005 x 1 / 1
= 0.005 moles of HCl :
M ( HCl ) = n / V
M ( HCl ) = 0.005 / 0.045
= 0.1111 M
hope this helps!
Answer:
d = 0.793 g/L
Explanation:
Given data:
Density of fluorine gas = ?
Pressure of gas = 0.554 atm
Temperature of gas = 50 °C (50+273.15K = 323.15 K)
Solution:
Formula:
PM = dRT
M = molar mass of gas
P = pressure
R = general gas constant
T = temperature
d = PM/RT
d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K
d = 21.05 atm.g/mol/26.53 atm.L /mol
d = 0.793 g/L
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
Learn more about limiting reagent here ;
brainly.com/question/11848702
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Answer: I think the right answer is c
