<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
<span>Answer: option B. 3.07 g
Explanation:
1) given reaction:
S(s) + O₂ (g) → SO(g)
2) Balanced chemical equation:
</span><span>2S(s) + O₂ (g) → 2SO(g)
3) Theoretical mole ratios:
2 mol S : 1 mol O₂ : 2 mol SO
3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa
use the ideal gas equation:
pV = nRT
with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)
=> n = pV / (RT) =
n = [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K) × 573.15 K ]
n = 0.0954 mol SO
4) proportion with the theoretical ratio S / SO
2 mol S x
-------------- = ----------------------
2 mol SO 0.0954 mol SO
=> x = 0.0954 mol S.
5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol
mass = number of moles × atomic mass
mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S
6) Therefore the answer is the option B. 3.07 g
</span>
Answer:
95.2 - 40.8 = 54.4 g of oxygen
number of moles = mass (g)/ Mr
no. of moles of carbon = 40.8/12 = 3.4
no. of moles of oxygen = 3.4
divide both by smallest value which is 3.4 and you’ll get 1 mole of carbon and 1 mole of oxygen therefore the empirical formula is CO
Explanation:
hope this helps :)
<u>Answer:</u>
<em>
needed to heat the amount of water giving.
</em>
<u>Explanation:</u>
To solve this question we first need to know the specific heat of water which is given by <em>4.18 joules</em> Per gram for increasing <em>1° of temperature</em>. Now since one gram of water name one joules per gram so 2 grams will need <em>twice of 4.18</em> that is 8.36 similarly 50 grams will need 
Now we need to calculate the heat for 10° of change because temperature rises from 50° centigrade to 60° centigrade and difference in degrees
<em>so heat needed is=
</em>
The (s) in the chemical name of (s)-3-iodo-2-methylnonane indicates an S-configuration using the Cahn-Ingold-Prelog system of stereochemical nomenclature. The S-configuration means that an "imaginary" rotation from the highest priority substituent group to the lowest priority substituent group of the chiral center moves counterclockwise (to the left), provided that the lowest priority group is oriented "towards the back" (symbolized by dashed lines).
The highest priority group (iodine in this case) is the one with the highest atomic number and the lowest priority (hydrogen in this case) is one with the lowest atomic number.
If the atoms directly beside the chiral center have the same atomic number (Carbon-2 and Carbon-4 in this case), the atoms next to them will be evaluated until a point of difference is found. Carbon-2 is connected to 2 other carbon atoms and 1 hydrogen atom, while Carbon-4 is connected to only 1 carbon atom and 2 hydrogen atoms. Thus, Carbon-2 has a higher priority, with the point of difference being the carbon atom of the methyl group attached to Carbon-2. Both Carbon-2 and Carbon-4 are connected to one carbon atom from the main nonane chain, but the other atoms connected to Carbon-4 are hydrogen atoms only. Carbon-2 has an extra carbon connected to it and carbon has a higher atomic number than hydrogen.
If there is no point of difference, the central atom is not chiral and cannot be named using the Cahn-Ingold-Prelog system.
Thus, the structure of (s)-3-iodo-2-methylnonane is