If you start with 40.0 grams of the element at noon, 10.0 grams
radioactive element will be left at 2 p.m. The correct answer between
all the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
Answer:
2.52 g NaCl
Explanation:
(Step 1)
To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:
Molarity = moles / volume (L)
After you convert mL to L, you can plug the given values into the equation and simplify to find moles.
136.9 mL / 1,000 = 0.1369 L
Molarity = moles / volume
0.315 M = moles / 0.1369 L
0.0431 = moles
(Step 2)
Now, you can use the molar mass to convert moles to grams.
Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol
Molar Mass (NaCl): 58.443 g/mol
0.0431 moles NaCl 58.443 g
------------------------------ x ------------------- = 2.52 g NaCl
1 mole
Mass of Na2SO4= 514.18 grams
<h3>Further explanation</h3>
Given
423.67 g of NaCl
Required
mass of Na2SO4
Solution
Reaction
2NaCl + H2SO4 → Na2SO4 + 2HCl
mol NaCl :
= 423.67 g : 58.5 g/mol
= 7.24
From the equation, mol Na2SO4 :
= 1/2 x mol NaCl
= 1/2 x 7.24
= 3.62
Mass Na2SO4 :
= 3.62 mol x 142,04 g/mol
= 514.18 grams
Answer: 40 grams
Explanation:
The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = 93.4J
M = ?
C = 0.129 J/g.C
Φ = 40.4°C - 22.3°C = 18.1°C
Then, Q = MCΦ
Make Mass, M the subject formula
M = Q/CΦ
M = (93.4J) / (0.129 J/g.C x 18.1°C)
M = 93.4J / 2.33J/g
M = 40 g
Thus, the mass of the lead is 40 grams
The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.
<h3>Gibb's free energy of reactions</h3>
It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;
On this note, it follows that;
Hence, the Gibb's free energy for the reaction is;
- ΔG = 14.6 - 71.52
- ΔG = -56.92KJ
Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.
Read more on Gibb's free energy;
brainly.com/question/13765848
