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.
<span>when the number of moles Ca = mass of Ca / molar mass of Ca.
and we can get the molar mass of Ca, it is = 40 g/mol
and we have already the mass of Ca (given) = 9.8 g
so, by substitution: the moles Ca = 9.8 g / 40 g/mol
= 0.245 moles</span>
Answer:
36.8 L
Explanation:
We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 80 °C
T(K) = 80 + 273
T(K) = 353 K
Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:
Number of mole (n) = 1.27 moles
Temperature (T) = 353 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 1.27 × 0.0821 × 353
V = 36.8 L
Thus, the volume occupied by the helium gas is 36.8 L
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>
m 9CH₃COOH: 60u×9 = <u>540u</u>
<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g
</u></span>
