Answer: The pH of the resulting solution will be 3.001
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
We are given:
Moles of NaOH = 0.0224 moles
Molarity of nitrous acid = 0.475 M
Molarity of sodium nitrite = 0.302 M
Volume of solution = 150 mL = 0.150 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]\text{Moles of nitrous acid}=(0.475mol/L\times 0.150L)=0.07125mol[/tex]
[tex]\text{Moles of sodium nitrite}=(0.302mol/L\times 0.150L)=0.0453mol[/tex]
The chemical equation for the reaction of nitrous acid and NaOH follows:
[tex]HNO_2+NaOH\rightleftharpoons NaNO_2+H_2O[/tex]
I: 0.07125 0.0224 0.0453
C: -0.0224 -0.0224 +0.0224
E: 0.04885 - 0.0677
The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:
[tex]pK_a=-\log K_a[/tex] ......(2)
We know:
[tex]K_a[/tex] for nitrous acid = [tex]7.2\times 10^{-4}[/tex]
Using equation 2:
[tex]pK_a=-\log (7.2\times 10^{-4})=3.143[/tex]
To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:
[tex]pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}[/tex] .......(3)
Given values:
[tex][NaNO_2]=\frac{0.0677}{0.150}[/tex]
[tex][HNO_2]=\frac{0.04885}{0.150}[/tex]
[tex]pK_a=3.143[/tex]
Putting values in equation 3. we get:
[tex]pH=3.143-\log \frac{(0.0677/0.150)}{(0.04885/0.150)}\\\\pH=3.143-0.142\\\\pH=3.001[/tex]
Hence, the pH of the resulting solution will be 3.001
A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?
Answer:
Mass of sample in mg = 15,285 mg
Explanation:
Given:
Volume of urine sample = 15 ml
Density of sample = 1.019 g/ml
FInd:
Mass of sample in mg
Computation:
Mass = density x volume
Mass of sample in mg = Volume of urine sample x Density of sample
Mass of sample in mg = 1.019 x 15
Mass of sample in mg = 15.285 gram
Mass of sample in mg = 15.285 x 1,000
Mass of sample in mg = 15,285 mg
Why does glucose and acentic acid have the same empirical formula
Answer:
Examples. Glucose (C6H12O6), ribose (C5H10O5), Acetic acid (C2H4O2), and formaldehyde (CH2O) all have different molecular formulas but the same empirical formula: CH2O.
Explanation:In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.
Naturally occurring gallium is a mixture of isotopes
that contains 90.11% of Ga-69 (atomic mass = 68.93
u) and 9.89% of Ga-71 (atomic mass 70.92 u).
What is the average atomic mass of naturally
occurring gallium?
A) 69.93 amu
C) 69.50 amu
B) 69.12 amu
D) 69.00 amu
You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution
Answer:
ee = 96%
Explanation:
Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the differences of amounts of enantiomers and the total amunt. that is:
ee = |A-B|/ A+B * 100
ee = |98%-2%| / 98+2 * 100
ee = 96%You are an intermediate product of an industrial process which intends to separate iron from its ore. A well known iron ore is hematite. Which of these ores does not contain iron?
Goethite
Malachite
Siderite
Limonite
Answer:
Malachite
Explanation:
Malachite is a copper carbonate hydroxide mineral, with the equation Cu2CO3(OH)2. This dark, green-joined mineral solidifies in the monoclinic precious stone framework, and frequently shapes botryoidal, sinewy, or stalagmitic masses, in cracks and profound, underground spaces, where the water table and aqueous liquids give the way to synthetic precipitation. So, the answer is malachite. Best of Luck!
A diver exhales a bubble with volume of 250 mL at pressure of 2.4 atm and temperature of 15 C. How many gas particulate in this bubble?
Answer:
1.5x10²² particulates
Explanation:
Assuming ideal behaviour, we can solve this problem by using the PV=nRT formula, where:
P = 2.4 atmV = 250 mL ⇒ 250 / 1000 = 0.250 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 15 °C ⇒ 15 + 273 = 288 KWe input the given data:
2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 KAnd solve for n:
n = 0.025 molFinally we calculate how many particulates are there in 0.025 moles, using Avogadro's number:
0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulatesHow much energy does an X-ray with an 8 nm (8 x 10-9m) wavelength have?
A. 1.99 x 10-25 J
B. 3.33 x 1016 J
C. 2.48 x 10-17 j
D. 8.28 x 10-26 J
Answer:
it would be option C
Explanation:
Speed of light = 3×10^8m/s
Planck's constant = 6.626×10^-34 Js
Wavelength = 8 x 10^-9 m
Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9
Energy = [19.878×10^(8-34)] / 8 x 10^-9
Energy = 2.48475 × 10^(-26+9)
Energy = 2.48×10^-17 J
Can someone please help with these 2?
Equilibrium shifts to the right.
OPTION A
Emily spills concentrated sodium hydroxide solution on her lab bench. What she should do first?
Answer:
Explanation: hell noo
1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 358 K and 1.20 atm.
4 FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)
Answer:
40.0L of SO2 are produced
Explanation:
To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:
Moles O2:
n = PV/RT
n = 1.20atm*55.0L / 0.082atmL/molK*358K
n = 2.25 moles of O2.
Clearly, limiting reactant is O2.
The moles of SO2 produced are:
2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2
Volume SO2:
V = nRT/P
V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm
V = 40.0L of SO2 are produced
An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.
How many milliliters of this suspension should be given to an infant weighing 13 lb.
Answer:
0.8853 mL
Explanation:
First we convert 13 lb to kg, keeping in mind that 1 lb = 0.454 kg:
13 lb * [tex]\frac{0.454kg}{1lb}[/tex] = 5.902 kgThen we calculate how many mg of acetaminophen should be given, using the recommended dose and infant mass:
15 mg/kg * 5.902 kg = 88.53 mgFinally we calculate the required mL of suspension, using its concentration:
88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mL