Answer:
47.36mL
Explanation:
Using Boyles law equation, which states that:
P1V1 = P2V2
Where;
V1 = initial volume (mL)
V2 = final volume (mL)
P1 = initial pressure (atm)
P2 = final pressure (atm)
Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm
First, we need to convert 152kPa to atm by dividing by 101
1kPa = 0.0099atm
152kPa = 1.505atm
P1V1 = P2V2
1.505 × 25.3 = 0.804 × V2
38.08 = 0.804V2
V2 = 38.08/0.804
V2 = 47.36mL
Answer:
34.9 g of Zn(OH)₂ is the maximum mass that can be formed
Explanation:
Let's state the reaction:
ZnO(s) + H₂O(l) → Zn(OH)₂ (aq)
First of all, we need to determine the moles of each reactant and state the limiting:
28.6 g . 1mol /81.38 g = 0.351 moles of ZnO
9.54 g . 1mol /18 g = 0.53 moles of water
As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.
Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:
0.351 mol . 99.4 g /1mol = 34.9 g
<span>H2CO3 <---> H+ + HCO3-
NaHCO3 <---> Na+ + HCO3-
When acid is added in the buffer, the excess H+ of that acid reacts with HCO3- to form H2CO3, and due to this NaHCO3 dissociates into HCO3- to attain the equilibrium. and hence there is no net effect of H+ due to pH remain almost constant.
when a base is added to the buffer, the OH- ion of base react eith H+ ion present in buffer, then to attain equilibrium of H+ ion, the H2CO3 dissociates to produce H+ ion, but now there is the excess of HCO3- due to which Na+ ion react with them to attain equilibrium of HCO3-. hence there is again no net change in H+ ion due to which pH remain constant.....</span>
Answer:
1000N is needed to be applied.
Explanation:
Machines make doing work easier. They allow us use small effort to carry out work on huge amount of load.
The mechanical advantage of a machine;
(M.A) =load/effort
M.A = 0.6
Load =600N
effort =?
0.6 = 600/effort
effort = 600/0.6
effort = 1000N
Answer:
The concentration of the dilute sample will be 0.361 g/ml
Explanation:
If a solution is diluted into 1:10 ratio then the amount of solute of that solution will be decreased by 10 times.
The initial concentration of the stock solution was 3.61g/ml but when the solution is diluted in 1:10 ratio the solute concentration is also decreased by 10 times.SO at present the solute concentration becomes 3.61/10=0.361 g/ml.
