Answer:
0.25 mol/L
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 4L
Initial concentration (C1) = 0.5 mol/L
Final volume (V2) = 4 + 4 = 8L
Final concentration (C2) =?
Applying the dilution formula, we can easily find the concentration of the diluted solution as follow:
C1V1 = C2V2
0.5 x 4 = C2 x 8
Divide both side by 8
C2 = (0.5 x 4 )/ 8
C2 = 0.25 mol/L
Therefore the concentration of the diluted solution is 0.25 mol/L
Answer:
16.9g of H₂O can be formed
Explanation:
Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
8.76g * (1mol / 2.016g) = 4.345 moles
<em>Moles O₂:</em>
PV = nRT
PV/RT = n
P = 1atm at STP
V = 10.5L
R = 0.082atmL/molK
T = 273.15K at STP
n = 1atm*10.5L / 0.082atmL/molK*273.15K
n = 0.469 moles of oxygen
For a complete reaction of 4.345 moles moles of hydrogen are required:
4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant
Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:
0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.
The mass is -Molar mas H₂O = 18.01g/mol-:
0.938 moles * (18.01g/mol) =
<h3>16.9g of H₂O can be formed</h3>
Answer:
- C. it is a strong acid since the blue litmus paper turns red and the pH is 1.7
Explanation:
As per table gastric juice has pH of 1.7 which is the strong acid. If you remember, 0-7 range determines acids and litmus paper confirms that by turning red.
So correct choice is C.
Answer:
A reaction rate is a measure of how fast a reactant disappears or a product forms during a reaction.
Explanation:
It is usually defined as the change in concentration per unit time:
Δ(concentration)/Δt
The units are (moles per litre) per second.
In symbols, the units are mol/(L·s) or mol·L^-1 s^-1.
Answer:
1.6 grams
Explanation:
We need to prepare 100 mL (0.100 L) of a 0.10 M CuSO₄ solution. The required moles of CuSO₄ are:
0.100 L × 0.10 mol/L = 0.010 mol
The molar mass of CuSO₄ is 159.61 g/mol. The mass corresponding to 0.010 moles is:
0.010 mol × (159.61 g/mol) = 1.6 g
We should use 1.6 grams of CuSO₄.
