The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is
it is reactive because it has to lose one electron to have a full outermost energy level.
<u><em>Explanation</em></u>
- <u><em> </em></u>Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.
- sodium has one electron in the outermost energy level.
- for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)
Answer:
3.2L
Explanation:
PV=nRT
since pressure and temperature are held constant we have V=nR
R is a constant also,
Thus; 
v1=1.5L , n1=3mol, n2=1.4mol
v2=
v2=3.2L
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
Let us differentiate accuracy from precision. Accuracy is the nearness of the measured value to the true or exact value. On the other hand, precision is the nearness of the measured values between each other. So, for precision, select the student in which the measured values are very near to each other. That would be Student III. Now, for accuracy, let's find the average for each student.
Student I: (<span>8.72g+8.74g+8.70g)/3 = 8.72 g
Student II: (</span><span>8.56g+8.77g+8.83g)/3 = 8.72 g
Student III: (</span><span>8.50g+8.48g+8.51g)/3 = 8.50 g
Student IV: (</span><span>8.41g+8.72g+8.55g)/3 = 8.56 g
From the given results, the accurate one would be Students I and II. So, we make a compromise. Even though Student III is precise, it is not accurate. If you compare between Students I and II, the more precise data would be Student I. Therefore, the answer is Student I.</span>
Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
<h3>The concentration of monosodium phosphate is 0.1262M</h3>
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