"<span>using shaving cream, rather than shaving on dry skin" I believe this is the answer. coz rest of them would increase friction rather than decrease</span>
Answer:
17.18 moles of NaCl are in 2,719 mL of a 6.32
Explanation:
Answer:
0.30 M
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t = ?
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 1.36 M
k is the rate constant = 0.208 s⁻¹
t = 7.30 seconds
So,
![[A_t]=1.36\times e^{-0.208\times 7.30}\ M=0.30\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D1.36%5Ctimes%20e%5E%7B-0.208%5Ctimes%207.30%7D%5C%20M%3D0.30%5C%20M)
Answer:
1.0 ° C
Explanation:
The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85
Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃
⇒ 17/85 = 1.38 moles
Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.
when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..
Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.
Using the relation : Q = mcΔT to determine the temperature drop ; we get:
55.2 = 17 × 4 (ΔT)
55.2 = 68 ΔT
ΔT= 0.8 ° C
ΔT ≅ 1.0 ° C
Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C
