Answer:
b. gelatin
Explanation:
a homogeneous noncrystalline consisting of large molecules or ultramicroscopic particles of one substance.
A solution is made by mixing of 42.g water and 77.g of acetic acid HCH3CO2. Calculate the mole fraction of water in this solution.
Moles of [tex]H_2O[/tex] ,
[tex]n_{H_2O}=\dfrac{42}{2\times 1 + 16}=\dfrac{42}{18}\\\\n_{H_2O}=2.33\ moles[/tex]
Moles of acetic acid [tex]HCH_3CO_2[/tex] ,
[tex]n_{A.A}=\dfrac{77}{1 + 12 + 3 + 12 + 16\times 2}=\dfrac{77}{60}\\\\n_{A.A}=1.28\ moles[/tex]
Mole fraction of water :
[tex]M.F_{H_2O}=\dfrac{n_{H_2O}}{n_{H_2O}+n_{A.A}}\\\\M.F_{H_2O}=\dfrac{2.33}{2.33+1.28}\\\\M.F_{H_2O}=0.645[/tex]
Therefore, mole fraction of water in this solution is 0.645 .
Hence, this is the required solution.
In Fig 1-2, how should the length indicated by the arrow along the ruler be recorded?
Answer:
C. 0.35cm
Explanation:
The length indicated by the arrow along the ruler should recorded be recorded as "0.35cm".
This is correct because when counting the measurement on the ruler, the first line on the ruler is 0.1cm, the second line is 0.2cm, and so on. The spaces between each line is 0.05cm. So, the arrow is pointing on the space between 0.3cm and 0.4cm.
Therefore, 0.3cm + 0.05cm = 0.35cm.(answer).
Fluorine is the first element in Group 17 of the periodic table. Bromine is the third element in the same group. Which statement is true of both
elements?
A. Both of their ions will have the same positive charge.
B. The charge on the ion formed by fluorine will be less than the charge on the bromine ion.
C. The charge on the ion formed by fluorine will be greater than the charge on the ion formed by bromine.
D. Both of their ions will have the same negative charge.
Answer:D. Both of their ions will have the same negative charge
Explanation:
what is the first step in the scientific inquiry process
Answer:
make an observation that describes a problem
Explanation:
Answer:
The first step in the Scientific Method is to make objective observations.
Explanation:
....-
Nicotinic acid, HC6H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4 × 10-5. Calculate [H+] and the pH of a 0.041 M solution of HC6H4NO2.
Answer:
[H+] = 7.576x10⁻⁴M
pH = 3.12
Explanation:
Based on the equilibrium of the nicotinic acid in water:
HC6H4NO2(aq) + H2O(l) ⇄ C6H4NO2-(aq) + H3O+(aq)
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
As both C6H4NO2-(aq) and H3O+(aq) comes from the same equilibrium, we can approximate their concentration as X and replace:
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
1.4x10⁻⁵ = [X] [X] / [0.041M]
5.47x10⁻⁷ = X²
7.576x10⁻⁴M = X = [H+]And as pH is defined as -log [H⁺]
pH = 3.12Which set of procedures would be best to follow when separating a mixture of sand, salt, and water?
Answer:
Pour the mixture through a filter to separate the sand, then evaporate the liquid to separate the salt.
Explanation:
There are various methods used to separate single substances from a mixture. The method adopted depends upon the type of mixture and the ingredients used in it. According to the given question, the mixture comprises of sand, salt and water. Here, the first process will be the process of filtration. Through filtration, the salt-water solution and sand can be separated. Salt-water solution can be separated by the process of evaporation in which sat and water will be separated individually.
3. A certain Chemical Industry company has a quality control job opening. The job is open for any major with basic knowledge of chemistry. You decided to apply. In the interview the HR personnel gives you a sealed folder from a certain lot to test your laboratory experience, as well as your quantitative and volumetric analysis skills. The chemical contained in the sample is benzoic acid (C-H602) and it is known to be a monoprotic acid. In order to get the job, you need to determine if the sample's purity is acceptable based on their standards. Inside the folder you found a vial with a solid sample labeled BA-I, a periodic table, and the following data: 1.250 g of the sample required 20.15 mL of 0.500 M concentration of NaOH to reach the end point. The lot can be denied if the purity is below 99.5 % purity.
1) What is the purity in the sample?
2) Is it the purity acceptable?
3) Would you repeat the titration experiment?
Answer:
1) 97.6%
2) No the purity is not acceptable because the standard is 99.5% purity.
3) Yes I will repeat the titration experiment to confirm my result.
Explanation:
Equation of the reaction;
C7H6O2(aq) + NaOH(aq) ---------> C7H5ONa(aq) + H2O(aq)
From the information provided;
Number of moles of NaOH reacted = concentration × volume = 20.15/1000 × 0.500 = 0.01 moles
From the reaction equation;
1 mole of C7H6O2 reacts with 1 mole of NaOH
Hence 0.01 moles of C7H6O2 will react with 0.01 moles of NaOH
Mass of C7H6O2 reacted = number of moles of C7H6O2 × molar mass of C7H6O2
Molar mass of C7H6O2 = 122.12 g/mol
Mass of C7H6O2 reacted = 0.01 moles × 122.12 g/mol = 1.22 g
Percentage by mass of pure C7H6O2 in the impure sample = 1.22/1.250 × 100 = 97.6 %
Into which two smaller groups are plants divided.
(a)vascular and nonvascular
(b)seed and seedless
(c)monocots and dicots
(d)seeds and spores
giving 17 points please help
4. Horizontal rows of the Periodic Table are called:
a, Clusters
Groups
b. Families
d) Periods
a. Describe the molecule chlorine dioxide, CIO in terms of three possible resonance structures.
b. Do any of these resonance structures satisfy the octet rule for every atom in the molecule? Why or why not?
Answer:
See explanation
Explanation:
The compound ClO2 has 19 valence electrons. ClO2 is a bent molecule with tetrahedral electron pair geometry but has two lone pairs of electrons. This is indicated by the presence of four electron pairs on the outermost shell of the central atom.
The molecule has an odd number of valence electrons, hence, it is generally regarded as a paramagnetic radical. None of the proposed Lewis structures for the molecule is satisfactory because none of them obeys the octet rule.
From the images attached, one can easily see that the electron dots around the oxygen and chlorine atoms does not satisfy the octet rule in all the resonance structures shown.
Observing a phenomenon in the lab includes which of the following?
Throwing away irrelevant items
Sorting through and disposing of police reports
Collecting evidence
Identifying unique features of evidence
Answer:
D
Explanation:
Identifying unique features of evidence
A laboratory is a place where scientific investigations and experiments are conducted. Observing a phenomenon in the lab includes identifying unique features of evidence. Thus, option D is correct.
What is observation?
An observation is said to be the deduction that is seen or heard and is due to the actions of the process or event that are gained from the information from a conducted experiment.
It is an important factor in an experiment and research and is used in the lab to falsify or prove the hypothesis. It is made based on the experimental setup and involves various steps.
The observations are used to draw conclusions and inferences based on the detailed identification of the characteristics of the evidence. The evidence supports the observation.
Therefore, option D. the features of evidence are the correct option.
Learn more about observations here:
https://brainly.com/question/28041973
#SPJ2
The Earth can gain or lose matter. True False
Answer:
True
Explanation:
According to some calculations, the Earth is losing 50,000 metric tons of mass every single year, even though an extra 40,000 metric tons of space dust converge onto the Earth's gravity well, it's still losing weight.
Answer:
true
Explanation:
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?
Answer:
3.13%.
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 0.15
Half life (t½) = 2 mins
Number of half-life (n) = 5
Fraction of sample remaining =.?
Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:
Amount remaining (N) = 1/2ⁿ × original amount (N₀)
NOTE: n is the number of half-life.
N = 1/2ⁿ × N₀
N = 1/2⁵ × 0.15
N = 1/32 × 0.15
N = 0.15/32
N = 4.69×10¯³
Therefore, 4.69×10¯³ is remaining after 5 half-life.
Finally, we shall the fraction of the sample remaining after 5 half-life as follow:
Original amount (N₀) = 0.15
Amount remaining (N) = 4.69×10¯³
Fraction remaining = N/N₀ × 100
Fraction remaining = 4.69×10¯³/0.15 × 100
Fraction remaining = 3.13%
What is the pH of a solution made by mixing 0.050 mol of NaCN with enough water to make a liter of solution
Answer:
pH = 11
Explanation:
The equilibrium of a weak base as NaCN in water is:
NaCN(aq) + H₂O(l) ⇄ OH⁻(aq) + Na⁺(aq) + HCN(aq)
And kb, the equilibrium constant, is:
Kb = [OH⁻] [HCN] / [NaCN]
Where Kb of NaCN is 2.04x10⁻⁵
In the beginning, the [NaCN] is 0.050mol / L = 0.050M.
Both [OH⁻] and [HCN] are produced from this equilibrium, and its concentration is X, that is:
2.04x10⁻⁵ = [X] [X] / [0.050M]
1.02x10⁻⁶ = X²
X = 1x10⁻³ = [OH⁻]
As pOH = - log [OH⁻]
pOH = 3.00
And pH = 14 - pOH
pH = 11
what element has a higher ionization energy carbon or silicon
Describe three factors that could limit the growth of the prairie dog population
Answer:
Changes in environment, food source changes and disease?
Explanation:
I dont know what the answer choices are
Answer:
changes in the environment
Explanation:
like digging big holes.
Hope this helps!
) What would be the effect on the molarity of the NaOH solution if some of the water evaporated from the Florence flask after the NaOH solution was standardized with the KHP
Answer:
The effect is the increasing of the molar concentration.
Explanation:
When you standarize a solution of NaOH with KHP you are establish its molar concentration (That is the amount of moles of NaOH per liter of solution).
If you evaporated some water of the solution, you are increasing its concentration because volume is decreasing doing the amount of moles per liter increasing.
which two options are examples of chemical changes?
A. A shinny metal bar glows red and expands when heated.
B. white piece of paper turns black and gives off a smell when burned
C. When two clear colorless liquids are mixed together, a white solid forms
D. A white powder mixed with a clear, colorless liquid makes a solution that is also clear, colorless liquid
Answer:
B
Explanation:
It's the only thing that actually changes into something different and when u burn it it makes chemicals
What specific portion of the gill is used to increase surface area though which gas (oxygen and carbon dioxide) exchange take place?
Answer:
gill lamellae
Explanation:
Gills in fishes are structures that permit fish to carry out gaseous exchange in water. Fishes exchange gases such as oxygen and carbon dioxide using gills.
Fish gills carry out their activity mainly through its major component called the gill lamellae. These are comb-like filaments which help increase the surface area of the gills in order to facilitate gaseous exchange
if u trust urself do it
Study these images.
4 photos of clouds. 1: Sky covered with large, flat layers of blue, grey clouds. 2: A tall, fluffy cloud shaped like an anvil. 3: Round, puffy clouds in a blue sky. 4: Thin, wispy clouds high in the sky.
Which image shows a cumulonimbus cloud?
1
2
3
Answer:
3
Explanation:
I wish you the best, its three or 2
Answer:
2
Explanation:
Edge 2021
Question 1 of 10
What is technology?
A. An understanding of something new.
B. The steps that engineers go through to create a product.
C. Something created using science for use by society.
D. A method that is used to solve problems,
SUBMIT
Answer:
C.
Explanation:
You can use the series of elimination for this. First, you look at A. Technology is not an understanding of something new, so we cross that out. Second, you look at B. Technology isn't a series of steps, so we can mark that one off. Third, you look at C. Technology is something created using science for use by society, so we can keep that in mind. Lastly, we check D. Technology doesn't match up to the definition, so we can cross that one out. The answer that would make the most sense would be D.
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic acid has a pKa of 3.75. 2. Create approximately 20 mL of buffer solution with a pH of 4.25.
Answer:
15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.
Explanation:
To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid
4.25 = 3.75 + log [A⁻] / [HA]
0.5 = log [A⁻] / [HA]
3.162 = [A⁻] / [HA] (1)
As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:
0.10M * 20x10⁻³L =
2x10⁻³moles = [A⁻] + [HA] (2)
Replacing (2) in (1):
3.162 = 2x10⁻³moles - [HA] / [HA]
3.162 [HA] = 2x10⁻³moles - [HA]
4.162[HA] = 2x10⁻³moles
[HA] = 4.805x10⁻⁴ moles
[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles
That means, to create the buffer you must add:
[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =
15.2mL of the 0.10M sodium formate solution[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =
4.8mL of the 0.10M formic acid solutionUse the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.
Answer:
The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].
The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].
Explanation:
The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.
Let [tex]\lambda_\text{vac}[/tex] denote the wavelength of the photon released when measured in vacuum.Let [tex]R_\text{H}[/tex] denote the Rydberg constant for hydrogen. [tex]R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}[/tex].Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the principal quantum number of the initial and final main energy level of that electron. (Both [tex]n_1\![/tex] and [tex]n_2\![/tex] should be positive integers; [tex]n_1 > n_2[/tex].)The Rydberg Equation gives the following relation:
[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].
Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:
[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].
In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:
[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].
Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].
Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:
[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].
Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].
Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:
[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].
The energy of the transition between n = 7 and n = 3 is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.
Using the Rydberg Equation for energy;
ΔE = -RH(1/n^2final - 1/n^2initial)
Given that;
nfinal = 3
ninitial = 7
RH = 2.18 × 10^-18 J
ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)
ΔE = - 2.18 × 10^-18(0.11 - 0.02)
ΔE = - 1.96 × 10^-19 J
For the second part;
Since the wavelength is 1000nm, we have;
λ = 1000nm
c = 3 × 10^8 m/s
f = ?
c = λf
f = c/λ
f = 3 × 10^8 m/s/1000 × 10^-9 m
f = 3 × 10^8 m/s/ 1 × 10^-6 m
f = 3 × 10^14 Hz
Learn more: https://brainly.com/question/18415575
2 2 6 2 6 2 10 3
1s 2s 2p 3s 3p 4s 3d 4p
=
Answer:
ARSENIC
Explanation:
It has an atomic number of 33
1. What 2 subatomic particles have charges? List the particle name and its charge.
Answer: Proton - positive charge (+)
Neutron - neutral charge (0)
Electron - negative charge (-)
Explanation:
A balloon contains 1.1 L of gas at a pressure of 0.80 atm. How will the volume
change if the pressure is increased to 2.0 atm?
Answer:
Final volume = 0.44 L
Explanation:
Given data:
Initial volume of balloon = 1.1 L
Initial pressure = 0.80 atm
Final volume = ?
Final pressure = 2.0 atm
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
0.80 atm × 1.1 L = 2.0 atm × V₂
V₂ = 0.88 atm. L/ 2.0 atm
V₂ = 0.44 L
Please help will give brainliest _________________________________________________________________________________________________________________________________________
Answer:
What subject is this for?
Explanation:
What happen when a piece of sodium is exposed in air?
Answer:
I literally got this from google.
Explanation:
In ordinary air, sodium metal reacts to form a sodium hydroxide film, which can rapidly absorb carbon dioxide from the air, forming sodium bicarbonate. ... In a comparatively dry atmosphere, sodium burns quietly, giving off a dense white caustic smoke, which can cause choking and coughing.
Happy almost Halloween! :)
A student measured the masses of four different-sized blocks. The student determined that each block had a mass of 50 grams.
(There is a small block, a little bit bigger block, a big block and the biggles block)
Which block has the least density?
Answer:..
Explanation:
The system is originally at equilibrium with [butane] = 1.0 M and [isobutane] = 2.5 M.If 0.50 mol/L of isobutane is suddenly added and the system shifts to a new equilibrium position, what is the equilibrium concentration of butane?
Answer:
1.14 M
Explanation:
Let's consider the following reaction.
butane ⇄ isobutane
We can use the concentrations at equilibrium to calculate the equilibrium constant.
Kc = [isobutane] / [butane]
Kc = 2.5 / 1.0
Kc = 2.5
If we add 0.50 M of isobutane, we get [isobutane] = 2.5 + 0.50 = 3.0 M.
This will be an initial concentration in an ICE chart.
butane ⇄ isobutane
I 1.0 3.0
C +x -x
E 1.0+x 3.0-x
The equilibrium constant is:
Kc = 2.5 = [isobutane] / [butane]
2.5 = (3.0-x) / (1.0+x)
2.5 + 2.5x = 3.0-x
x = 0.14
The equilibrium concentration of butane is:
[butane] = 1.0+x = 1.14 M