Answer: The gray mice would immediately be eaten by the predators in the area and go extinct.
Explanation:
Natural selection is the differential survival of some members of the population of species which have better phenotypic advantage over others. Such members will survive and reproduce. According to the given situation, white mice were the choice of food for predators and the recessive trait of gray color of gray mice was favoring their survival as they may remain undetected by the predators. But the predators were able to detect the gray mice and changed their food preference from white to gray this way predators will consume all the gray mice and they may extinct in future as it is not possible to change the color of the fur from white to gray so quickly. The adaptation and passage of white trait among the members of the population will occur with a gradual change in successive generations of the population.
On analysis of 3.4 mg of an organic compound, 8.03 g of CO2 and 3.34 g of H2O were obtained. The percentage composition of carbon and hydrogen in the compound are.......................... respectively
Answer: The percentage composition of carbon and hydrogen in the compound are 64.4 % and 10.9 % respectively
Explanation:
Mass of [tex]CO_2[/tex] = 8.03 g
Mass of [tex]H_2O[/tex]= 3.34 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 8.03 g of carbon dioxide, =[tex]\frac{12}{44}\times 8.03=2.19g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 3.34 g of water, =[tex]\frac{2}{18}\times 3.34=0.371g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (3.4) - (2.19+0.371) = 0.839 g
percent composition of C =[tex] \frac{\text{mass of C}}{\text{ Total mass}}\times 100= \frac{2.19}{3.4}\times 1000=64.4\%[/tex]
percent composition of H =[tex] \frac{\text{mass of H}}{\text{ Total mass}}\times 100= \frac{0.371}{3.4}\times 1000=10.9\%[/tex]
Thus the percentage composition of carbon and hydrogen in the compound are 64.4 % and 10.9 % respectively
What is the law of conservation or matter? Please helppppppp.....no links no trolling otherwise I'll report!
May give brainliest to first correct. :)
Answer:
In physics, a conservation law states that a particular measurable property of an isolated physical system does not change as the system evolves over time. Exact conservation laws include conservation of energy, conservation of linear momentum, conservation of angular momentum, and conservation of electric charge.
Explanation:
hope this helps :)
Which of the following reactions best describes the acid-base properties of acetic acid (CH3COOH) in an aqueous solution? O CH3COOH(aq) + 4H2O(l) 4H30+ (aq) + C2024-(aq) (reversible reaction) O CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq) (reversible reaction) O CH3COOH(aq) + H2O(1) CH3C00* (aq) + OH(aq) (reversible reaction) -> H2O*(aq) + O CH3COOH(aq) + H2O() CH3COO-(aq) O CH3COOH(aq) + H2O(0) + OH(aq) -> CH3COO(aq)
Answer:
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
Explanation:
Acetic acid is a weak acid, so it only partially dissociates, hence the reversible reaction. Acids react with water to produce conjugate base and hydronium.
Write down at least one example of ketone with its IUPAC name
Answer:Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone.
Explanation:
Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. ????????????H(????) → ????????+(????????) + ????H −(????????) + x1????????
Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride. ????????????H(????) + H +(????????) + ????????−(????????) → H2????(????) + ????????+(????????) + ????????−(????????) + x2????????
Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride. ????????+(????????) + ????H −(????????) + H +(????????) + ????????−(????????) → H2????(????) + ????????+(????????) + ????????−(????????) + x3????J
Procedure Reaction
1 a. In the glassware menu, take out a 50 mL graduated cylinder and a foam cup. From the tools menu, take out the scale. From the solutions stockroom, move the distilled water and solid NaOH onto the workbench.
b. Transfer 50.0 mL of water to the foam cup. To do this, drag the carboy of water onto the graduated cylinder. (Before you release the mouse button, the cursor will show a "plus sign" to indicate that it is the recipient). A transfer textbar will appear, enter "50.0" mL and click on pour. (You will notice that the graduated cylinder now reads 50.0 mL).
c. Weigh about 1 gram of solid sodium hydroxide pellets, NaOH(s), directly into the foam cup and record its mass to the nearest 0.01 gram. To do this, place the foam cup on the balance so it registers a mass, then click the "Tare" button. Drag the NaOH bottle onto the foam cup. (When you release the mouse, the bottle will be tipped to show that it is in the pour mode). Next, type "1.00" grams into the transfer bar and then click pour. Note that the balance now reads the mass of the transferred NaOH. You may now take the cup off of the scale.
d. Click on the graduated cylinder, record its temperature and then drag it onto the foam cup. (When you release the mouse, the graduated cylinder will be tipped to show that it is in pour mode.) Enter "50.0" mL in the transfer bar and then click pour. Record the highest temperature. e. Remove the foam cup and graduated cylinder from the workbench. (Right click on the item and select "remove.")
Reaction 2
a. Take the 0.5 M HCl from the strong acids cabinet and a fresh foam cup and a fresh 50 mL graduated cylinder from the glassware menu and place them on the workbench. The procedure for Reaction 2 is the same as for Reaction 1 except that 50.0 mL of 0.50 M hydrochloric acid solution is used in place of the water. After measuring 50.0 mL of the HCl solution into the graduated cylinder, proceed as before with steps b-e of the procedure for Reaction 1.
Reaction 3
a. Take out a 25 mL graduated cylinder, a fresh foam cup, the 1.0 M HCl and the 1.0 M NaOH. (If you are running out of room on the workbench, you may remove the previously used chemicals.) Use the graduated cylinder to measure and transfer 25.0 mL of 1.0 M HCl into the foam cup. Pour an equal volume of 1.0 M sodium hydroxide solution into a clean graduated cylinder.
b. Record the temperature of each solution to the nearest 0.1 oC. Pour the sodium hydroxide solution into the foam cup and record the highest temperature obtained during the reaction.
Data and Analysis
Reaction 1Reaction 2Reaction 3
Mass of solution* (g) 1.03g 1.03g
Initial temperature(°C) 25oC 25OC 25OC
Maximum temperature (°C) 30.3oC 37oC 31.7oC
Temperature change (∆T)
Heat energy q (kJ)
Moles of NaOH
Molar heat of reaction (-q/mol) also known as Enthalpy change,
DH (kJ/mol)
The conversion of more than one substance reactant into one or more distinct substances, products, and subsequent discussion can be characterized as follows:
Reaction Calculation:Calculating the Reaction 1:
[tex]NaOH\ (s) \rightarrow Na^+ \ (aq) + OH^- \ (aq) + X_1\ \ KJ ......................... (1)[/tex]
[tex]NaOH[/tex] mass = [tex]1\ g[/tex]
[tex]H_2O[/tex] mass = [tex]50 \ mL = 50\ g[/tex]
water heat of [tex]s_p[/tex] = [tex]4.186\ \frac{ J}{ g\ ^{\circ}C}[/tex]
[tex]\Delta T[/tex] = final temp - initial temp [tex]= 30.3 - 25 = 5.3^{\circ} \ C\\[/tex]
Therefore
Calculating the releasing heat
= mass × sp heat × [tex]\Delta T[/tex]
= 50 × 4.186 × 5.3 J
= 1109.3 J
Calculating the [tex]NaOH[/tex] mass [tex]= 1\ g = \frac{1}{ 40}\ mole= 0.025 \ mole[/tex]
Calculating the releasing heat per mole:
[tex]\to NaOH = \frac{1109.3}{ 0.025} = 44372\ J = 44.4\ KJ[/tex]
Thus
[tex]\to X_1 = 44.4\ KJ[/tex]
Calculating the Reaction 2:
[tex]NaOH \ (s) + H^+\ (aq) + Cl^- \ (aq) \rightarrow Na^+ \ (aq) + Cl^- \ (aq) + H_2O + X_2 \ KJ\\[/tex]
Calculating the net ionic from the equation:
[tex]NaOH\ (s) + H^+\ (aq) \rightarrow Na^+ \ (aq) + H_2O \ (l) + X_2 \ KJ ................................... (2)[/tex]
Calculating the [tex]NaOH[/tex] mass:
[tex]= 1\ g = \frac{1 }{ 40} = 0.025\ mole[/tex]
Calculating the [tex]HCl[/tex] mass:
[tex]= 50\ mL = 50\ g[/tex] [ density = 1 approx]
sp heat of the solution [tex]= 4.186 \frac{J}{g\ ^{\circ}C}[/tex] [ assume the sp heat same as water]
[tex]\Delta T[/tex] = final temp - initial temp [tex]= 36.97 - 25 = 11.97^{\circ} \ C[/tex]
Calculating the releasing heat:
= mass × sp heat × [tex]\Delta T[/tex]
= 50 × 4.186 × 11.97 J
= 2505.3 J
Calculating the releasing heat per mole in [tex]NaOH[/tex]:
[tex]= \frac{ 2505.3 }{ 0.025} = 100212\ J = 100.2 KJ[/tex]
Thus
[tex]X_2 = 100.2 \ KJ[/tex]
Calculating the Reaction 3:
[tex]Na^+ \ (aq) + OH^-\ (aq) + H^+ \ (aq) + Cl^- \ (aq) \rightarrow Na^+\ (aq) + Cl^-\ (aq) + H_2O + X_3\ KJ[/tex]
Calculating the net ionic in the given equation
[tex]H^+ + OH\rightarrow H_2O\ (l) + X_3\ KJ .............................................................. (3)[/tex]
Calculating the volume of [tex]NaOH[/tex]:
[tex]= 25 \ mL\ of\ 1.0\ M = 25 \times \frac{1 }{ 1000} \ mole = 0.025 \ mole[/tex]
Calculating the volume of HCl:
[tex]= 25 \ mL\ of\ 1.0\ M = 25 \times \frac{1 }{ 1000} \ mole = 0.025 \ mole[/tex]
Calculating the total volume
[tex]= 50 \ mL = 50\ g[/tex] { density = 1]
Calculating the sp heat in the solution
[tex]= 4.186 \frac{J}{ g \ ^{\circ} C}[/tex] [ assumed the sp heat is the same as water]
[tex]\Delta T[/tex] = final temp - initial temp [tex]= 31.7- 25 = 6.7^{\circ}\ C[/tex]
Calculating the releasing heat
= mass × sp heat × [tex]\Delta T[/tex]
= 50 × 4.186 × 6.7 J
= 1402.3 J
Calculating the releasing heat per mole in [tex]NaOH[/tex]:
[tex]=\frac{1402.3 }{ 0.025} \ J\\\\= 56092\ J\\\\= 56,09\ KJ[/tex]
Therefore
[tex]X_3 = 56.09 \ KJ\\\\X_1 = 44.4\ KJ\\\\X_2 = 100.2\ KJ\\\\X_3 = 56.09\ KJ\\\\X_2 - [ X_1+ X_3 ] = 100.2 - [44.4 + 56.09]\ = 100.2 - 100.49= -0.29[/tex]
So, the difference is equal to zero.
[tex]\to X_2 = X_1 + X_3[/tex]
This is due to the fact that if we add the reaction (1) and (3) we get the reaction (2)
Calculating the difference percentage:
[tex]= [\frac{0.29 }{100.2} ] \times 100 = 0.29\%[/tex]
The number of joules released in reaction 1 would be 4 times what is released in the calculation if we used 4 g of [tex]NaOH[/tex].
[tex]\to 4 \times 1109.3\ J = 4437.2 \ J\\\\[/tex]
Calculating the [tex]NaOH[/tex] moles [tex]= \frac{4}{40} = 0.1[/tex]
[tex]\to X_1 = \frac{4437.2}{ 0.1} = 44372 \ J = 44.4\ KJ[/tex]
As a result, it has no bearing on the solution's molar heat.
Find out more about the reaction here:
brainly.com/question/17434463
1. Chlorofluorocarbons (CFCs) A carbon-chlorine bond in the CFC molecule can be broken by sunlight, leaving a highly reactive free radical which then goes on to destroy the surrounding ozone molecules. The energy of a C-Cl bond is 328 kJ/mole. Calculate the wavelength of light needed to break a bond in a single molecule. In which region of the spectrum (infrared, visible, UV) does this wavelength fall
Answer: The wavelength for this photon is 365 nm. The wavelength corresponds to UV region.
Explanation:
The relationship between wavelength and energy of the wave follows the equation:
[tex]E=\frac{Nhc}{\lambda}[/tex]
E= energy
N = avogadros number
[tex]\lambda [/tex] = wavelength of the wave
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]328\times 10^3J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8m/s}{\lambda}[/tex]
[tex]\lambda=3.65\times 10^{-7}m=365nm[/tex] [tex]1m=10^9nm[/tex]
Thus wavelength for this photon is 365 nm. The wavelength of 365 nm corresponds to UV region.
How many grams of MnCl2 need to be added to 5 liters of water to make a 0.25 M solution?
Answer:
I think it is 860 g MG Cl2
Explanation:
your welcome
What must products equal in a chemical equation
Answer:
Reactants
Explanation:
In a chemical reaction, reactants must react to form products, so products are equal to the reactants
Vines most successfully compete for sunlight in tropical rainforests by _______.
a.
climbing up into the canopy
b.
strangling the trees they grow on
c.
having broader leaves that absorb more light
d.
having thick, woody stems
Answer:
climbing up into the canopy
Explanation:
I have an experiment on a mouse, and this experiment includes arsenic ... but I don't know where to buy it. Please help me. What are the characteristics of arsenic? Please help
Answer:
Arsenic is a semi-metal. In its metallic form it is bright, silver-grey and brittle. Arsenic is a well-known poison. Arsenic compounds are sometimes used as rat poisons and insecticides but their use is strictly controlled.
Explanation:
What product(s) would be formed when these are the reactants? C5H12 + O2 (limited)
Answer:
Carbon dioxide and water I believe because it is a combustion reaction
(a)0.444 Mol of COCl2 in 0.654 L of solution
(b) 98.0 gram of phosphoric acid, H3PO4, in 100 L of solution
(c) 0.2074 g of calcium hydroxide, Ca(OH)2 in 40.00 mL of solution
(d)10.5 kg of Na2AO4.10H2O in 18.60 L of solution
(e) 7.0 x 10-3 Mol of l2 in 100.0 ml of solution
(f) 1.8 x 104 mg of HCL in 0.075 of solution
Answer is in a photo. I can only upload it to a file hosting service. link below!
bit.[tex]^{}[/tex]ly/3a8Nt8n
You have a solution of NaOH that is 3.0 M. If you need 250 mL of a 1.5 M solution, what volume of the stock solution should you use?
Answer:
answer is 30 g!!!!!!!!!
take it before its gone
copper chloride reacts with magnesium but magnesium chloride does not react with copper. Why is this?
Answer:
Mg is highly reactive as compared to that of Copper owing to its periodic position
Explanation:
MG lies in the second column of the periodic table while Cu lies in the 11th column of the periodic table from the left.
The elements in the first two columns are metallic in nature are highly reactive.
In CuCl2 solution, Mg being highly reactive replaces Cu but in MgCl2 solution, Cu being less reactive in nature do not replaces the Mg from the solution.
how many liters are present in 31.998 g of O2?
Answer:
Explanation:
First calculate how many moles of O2 you have. O is 16g/mole, so O2 is 32g/mole. 50/32 = 1.5625 moles. 1 mole of any gas at stp is 22.4 liters.
1.5625 × 22.4 = 35 liters.
Its been 15 years since I graduated HS, and I still remember how to do this. Do your own homework next time and in 15 years you may be able to do the same.
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Calculate the number of moles of NO in 4.10 x 1021 molecules of NO.
Answer:
.0068
Explanation:
You divided the number of molecules by 6.022*10^23 (Avagodros Number) to get the number of moles.
What mass of Na2SO4 (sodium sulfate) is needed to make 2.5 L of 2.OM
solution?
O 1789
O 2849
O 7109
O 356 g
what is a derived lipid
Answer: Derived lipids:
Hydrolysis product of simple and compound lipids is called derived lipids. They include fatty acid, glycerol, sphingosine and steroid derivatives. Steroid derivatives are phenanthrene structures that are quite different from lipids made up of fatty acids.
Explanation:
What is the reaction order with respect to [A]
Answer:
it is the 1st order
Explanation:
i took that one and its the first order
which of the following is a good conductor of electricity?
iron
pure water
paper
wood
please help!!!!!
Answer:
Explanation:
Iron is the correct answer
Copper metal reacts with nitric acid according the equation below 3 Cu (s) +8 HNO 3(aq) -...m+3 Cu^ 2+ (aq)+2NO (g) +6 NO 3 (aq)+4 H 2 O 0 What mass of NO (g) will be produced if 8.96 g of copper are reacted wtih excess nitric acid. Be sure to include units.
Answer:
i think it is NO2 and +6 NO 6 that's all I have
A gas sample has a volume of 185 mL at 45°C. What is its volume at 109°C?
Answer:
342 mL
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 185 mL
Initial temperature (T₁) = 45 °C
Final temperature (T₂) = 109 °C
Final volume (V₂) =?
Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 45 °C
Initial temperature (T₁) = 45 °C + 273
Initial temperature (T₁) = 318 K
Final temperature (T₂) = 109 °C
Final temperature (T₂) = 109 °C + 273
Final temperature (T₂) = 382 K
Finally, we shall determine the final volume of the gas as follow:
Initial volume (V₁) = 185 mL
Initial temperature (T₁) = 318 K
Final temperature (T₂) = 382 K
Final volume (V₂) =?
V₁/T₁ = V₂/T₂
285 / 318 = V₂ / 382
Cross multiply
318 × V₂ = 285 × 382
318 × V₂ = 108870
Divide both side by 318
V₂ = 108870 / 318
V₂ ≈ 342 mL
Therefore, the new volume of the gas is 342 mL
Answer:
342 mL
Explanation:
Hope this helps
Define What is boiling
Answer:
Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the temperature at which the vapour pressure of the liquid is equal to the pressure exerted on the liquid by the surrounding atmosphere.
Explanation:
Hi
Below is a diagram of a man pushing a furniture to the left with 100N of force. Which of the following statements is incorrect?
A.) The furniture does not move if the force of friction is 100N to the right.
B.) The force exerted by the man on the furniture is equal to the force exerted by the furniture on the man.
C.) The force of friction must be lower than 100N for the cabinet to move.
D.) The force exerted by the man and the force of friction must be balanced in order to accelerate it to the left.
Answer:
If I'm not mistake it should be D
Explanation:
Please inform me if im right or wrong.
A buffer solution contains 0.20 mol of propionic acid (CH3CH2COOH) and 0.25 mol of sodium propionate (CH3CH2COONa) in 1.50 dm3.
What is the pH of this buffer?
Enter your answer using two decimal places.
Answer:
I don't know how to do it the subject
Please help I will give you brainly
Observe the activity in your classroom. You can observe people, objects, and parts of your actual classroom as you
begin to notice changes in matter. Collect data for two examples of changes in matter. Use the Changes in Matter
observation questions to determine if your examples are physical or chemical changes.
Your answer should include the following: two examples, the type of change for each, and how the examples were
formed. You can use this format:
Example 1
Type of Change:
How examples were formed:
Please help me
Answer: Microphones are a type of transducer - a device which converts energy from one form to another. Microphones convert acoustical energy (sound waves) into electrical energy (the audio signal). Different types of microphone have different ways of converting energy but they all share one thing in common! Furthermore, Example 1: Crushing a can.
Type of change:
Developmental – May be either planned or emergent; it is first order, or incremental. ...
Transitional – Seeks to achieve a known desired state that is different from the existing one. ...
Transformational – Is radical or second order in nature.
Determine whether each described process is endothermic or exothermic.
Wood burns in a fireplace Choose...
Ice melts into liquid water Choose...
Solid dissolves into solution, making ice pack feel cold Choose...
A process with a calculated positive q Choose...
A process with a calculated negative q Choose...
Acid and base are mixed, making test tube feel hot Choose...
Answer:
Following are the responses to the given choices:
Explanation:
Air woods is a smoking process as it releases heat.Incasereaction produces reaction energy, the response is then exothermic, while absorbs react energy, therefore the response is exothermic.Heat is essential for melting ice. Correspondingly, ice melts into liquid water as well as other reactions stop.Feel cold due to fuel absorption. That ice pack thus feels cold and brings another micro reaction to the stop.Unless the reaction's heat is positive, that process is endothermal.The reaction is exothermic unless the heat from the reaction is bad.Feel hot due to its loosening energy. A test tubefeelhot is, thus, an exothermic reaction.Which of the following are in our solar system? Select all answers that apply.
A the star Canis Majoris
B Jupiter's Moons
C Andromeda Galaxy
D Earth
E Asteroid Belt
Answer:
hola como esats UwU
Explanation:
yo bien y tu? UwU
Dust and particulates released in the home may cause
Lung cancer
Increased allergies and respiratory problems
Nausea
Answer:
b) increased allergies and respiratory problems.
Explanation:
i hope this helps!
have a great rest of your day!
please consider giving brainliest :)
Answer:
Human health effects of dust relate mainly to the size of dust particles. Dust may contain microscopic solids or liquid droplets that are small enough to get deep into the lungs and cause serious health problems. Large particles may irritate the nose, throat and eyes. ... Dust from soil can irritate the respiratory tract.
b. increased allergies and respiratory problems