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
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
Answer:
1.03 M
Explanation:
Step 1: Write the balanced equation
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
30.0 mL (0.0300 L) of 0.500 M HCl react.
0.0300 L × 0.500 mol/L = 0.0150 mol
Step 3: Calculate the moles of NaOH that react with 0.0150 moles of HCl
The molar ratio of NaOH to HCl is 1:1. The moles of NaOH that react are 1/1 × 0.0150 mol = 0.0150 mol.
Step 4: Calculate the molar concentration of NaOH
0.0150 moles of NaOH are in 14.5 mL (0.0145 L).
M = 0.0150 mol/0.0145 L = 1.03 M
Answer:
3
Explanation:
Applying,
= R/R'............... Equation 1
Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.
From the question,
Given: R = 40 atoms, R' = 5 atoms
Substitute these values into equation 1
= 40/5
= 8
= 2³
Equation the base,
n' = 3
Answer:
4.5 g/L.
Explanation:
- To solve this problem, we must mention Henry's law.
- Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
- It can be expressed as: P = KS,
P is the partial pressure of the gas above the solution.
K is the Henry's law constant,
S is the solubility of the gas.
- At two different pressures, we have two different solubilities of the gas.
<em>∴ P₁S₂ = P₂S₁.</em>
P₁ = 525.0 kPa & S₁ = 10.5 g/L.
P₂ = 225.0 kPa & S₂ = ??? g/L.
∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.