Answer: 0.4533mol/L
Explanation:
Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol
68g of CaCO3 dissolves in 1.5L of solution.
Xg of CaCO3 will dissolve in 1L i.e
Xg of CaCO3 = 68/1.5 = 45.33g/L
Molarity = Mass conc.(g/L) / molar Mass
Molarity = 45.33/100 = 0.4533mol/L
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
Answer: 7
Explanation:
Before a number but after a decimal. The zeros at the end would usually mean that it doesn't count but since the numbers are before the zeros and after a decimal it's 7 sig figs
I just took a test with this question and got the answer wrong for saying ethane. The correct answer is propane.
