Answer:
c) HOCH2CH2CH2OH
Explanation:
Hope this helps
What type of consumer eats only producers?
Answer: Primary consumers make up the second trophic level. They are also called herbivores. They eat primary producers—plants or algae—and nothing else. For example, a grasshopper living in the Everglades is a primary consumer
Answer:
primary consumer make up the second trophic level .they are also herbivores they eat primary consumer plants or alger and nothing else .for example a grasshopper living in the everglades is a primary consumer
After a metal object turns to dust (rust), it would have _____.
Need Help please
Answer:
The answer is high entropy.
Explanation:
Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCl (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCl(g) → H2(g) + Cl2(g)
AH = +184.6 kJ
2H2(g) + 12 N2(g) + 2 Cl2(g) NH4Cl(s) AH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
AH = +184.6 kJ
Answer:
(we use hess's law) it is so simple but the second reaction is not correct please right it
I need help pls!!!!
Answer:
A
Explanation:
jsgsnsbnss care about explaimi
Which of the following behaviors might indicate a patient is drug seeking?
A. A patient wants to avoid a specific medication because of
potential side effects.
B. A patient explains that she is from out of town and needs a
specific medicine because she left her prescription at home.
C. A patient fears her new prescription will conflict with another
medication she's currently taking.
D. A patient thinks she needs a smaller dose of her prescription
because it gives her headaches.
SUBMIT
Answer:
B is correct :)
Explanation:
Trust me I just took the test
What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?
Pls and thank you :) !
Answer:
36.8 L
Explanation:
We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 80 °C
T(K) = 80 + 273
T(K) = 353 K
Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:
Number of mole (n) = 1.27 moles
Temperature (T) = 353 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 1.27 × 0.0821 × 353
V = 36.8 L
Thus, the volume occupied by the helium gas is 36.8 L
help me plss plss.plsss.plsssplss plsss.plsss plsss plsss plsss
Answer:
1. Distance = 5 cm
Displacement = 1 cm
2. Distance = 7 cm
Displacement = 1 cm
3. Distance = 5 cm
Displacement = -1 cm
4. Distance = 6 cm
Displacement = -2 cm
5. Distance = 4 cm
Displacement = 0 cm
Explanation:
The distance an object travels is a measure of the total ground cobpvered bybthe object without regard to the direction of motion. It is a scalar quantity as it only takes into account magnitude but not the direction of a quantity.
Displacement however, is a measure of how far away or apart an object is from its starting position. It a vector quantity as it takes into account both the magnitude and direction in its measurement.
From the attachment provided, considering forward direction as positive and reverse as negative :
1. When a object moves from point 1 to point 4 and reverses to point 2.:
Distance = from 1 to 4 = 3 cm; from 4 back to 2 = 2 cm; 3 + 2 = 5 cm
Displacement = from 1 to 4 = +3 cm; from 4 back to 2 = -2 cm; 3 - 2 = 1 cm
2. When an object moves from point 1 to 5 and reverses to 2:
Distance = from 1 to 5 = 4cm; from 5 back to 2 = 3 cm; 4 + 3 = 7 cm
Displacement = from 1 to 5 = +4 cm; from 5 back to 2 = -4 cm; 4 - 3 = 1 cm
3. When an object moves from point 1 to 3 and then reverses to 0.
Distance = from 1 to 3 = 2 cm; from 3 back to 0 = 3 cm; 2 + 3 = 5 cm
Displacement = from 1 to 3 = +2 cm; from 3 back to 0 = -3 cm; 2 - 3 = -1 cm
4. When an object moves from point 3 to 5 and then reverses to 1:
Distance = from 3 to 5 = 2 cm; from 5 back to 1 = 4 cm; 2 + 4 = 6 cm
Displacement = from 3 to 5 = +2 cm; from 5 back to 1 = -4 cm; 2 - 4 = -2 cm
5. When an object moves from point 2 to 4 and then reverses to 3.
Distance = from 2 to 4 = 2 cm; from 4 back to 2 = 2 cm; 2 + 2 = 4 cm
Displacement = from 2 to 4 = +2 cm; from 4 back to 2 = -2 cm; 2 - 2 = 0 cm
What is the greatest concentration of pollution
Answer:
B
Explanation:The poisonous substances, present in the environment can easily get into the trophic level as living organism depends on each other and environment for food and nutrition. These poisonous substances may not be broken down in the body or excreted easily, efficiently and quickly. Instead, they accumulate in the tissues, and as the living organism eats more, the concentration of these substances increases and they pass from one trophic level to the next. The tertiary consumer being at the top of trophic levels receives the maximum pollutant. This phenomenon is known as biological magnification.
So, the correct answer is option B.
6.3x10-5M. What is the pH?
Answer:
https://sensorex.com/ph-calculator/
Explanation:
go to this link and then scroll down a little bit and then you'll find the answer
Consider the following ionization reaction.
HNO3 (aq) + H20 () = NO,- (aq) + H2O* (aq)
a)
Define acid and base according to Bronsted-Lowry, and give an example of each.
(3 marks)
b) Identify all acids and its conjugate base pairs from the above equation.
(2 marks)
Answer:
See explanation
Explanation:
According to Bronsted-Lowry, an acid is a proton donor while a base is a proton acceptor.
Hence, if we consider the reaction above, we will notice that for each base there is a conjugate acid and for each acid there is a conjugate base.
For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.
An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 435 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.
Answer:
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
Explanation:
From the question we are told that
Light wavelength [tex]\lambda_l=655nm[/tex]
Light wavelength atom fall [tex]x_L=5th-2nd[/tex]
Photon wavelength [tex]\lambda_p=435nm[/tex]
Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]
Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by
[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]
Therefore for initial drop of 5th to 2nd
[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]
Therefore for initial drop of 6th to 2nd
[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
Generally we subtract (5th to 2nd) from (6th to 2nd)
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]
Therefore for 6th to 5th stage is mathematically given by
[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]
[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
Write balanced equations for all the reactions in the catabolism of glucose to two molecules of glyceraldehyde 3-phosphate (the preparatory phase of glycolysis), including the standard free-energy change for each reaction. Then write the overall or net equation for the preparatory phase of glycolysis, with the net standard free-energy change.
Solution :
The balanced chemical equation are
The Catabolism of the glucose takes place in five stages :
1. Glucose + [tex]$ATP$[/tex] → glucose - [tex]$6$[/tex] - phosphate [tex]$+ ADP$[/tex] , [tex]$\Delta G^0=-16.7 \ kJ/mol$[/tex]
2. Glucose - [tex]$6$[/tex] - phosphate → Fructose - [tex]$6$[/tex] - phosphate , [tex]$\Delta G^0=1.7 \ kJ/mol$[/tex]
3. Glucose - [tex]$6$[/tex] - phosphate + [tex]$ATP$[/tex] → [tex]$ADP$[/tex] + Fructose - [tex]$1,6- \text{biophosphate}$[/tex], [tex]$\Delta G^0=-14.2 \ kJ/mol$[/tex]
4. Fructose - [tex]$1,6- \text{biophosphate}$[/tex] → dihydroxyacetonephosphate + glyceraldehyde [tex]$-3 - $[/tex] phosphate, [tex]$\Delta G^0=23.8 \ kJ/mol$[/tex]
5. Dihydroxyacetonephosphate → glyceraldehyde [tex]$-3 - $[/tex] phosphate , [tex]$\Delta G^0=7.5 \ kJ/mol$[/tex]
Therefore, the overall net equations
Glucose + [tex]$2ATP \rightarrow$[/tex] glyceraldehyde [tex]$-3 - $[/tex] phosphate + [tex]$2ADP$[/tex] [tex]$\Delta G^0=2.1 \ kJ/mol$[/tex]
atoms that have a high electronegativity...
Helium, bromine, agron, neon,flourine
(a) Using the expression Ka=[H+][A−][HA] , explain how to determine which solution has the lower pH , 0.10MHF(aq) or 0.10MHC2H3O2(aq) . Do not perform any numerical calculations.
Answer:
Ka=[H+][A−]/[HA]
which solution has the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq) :
Higher value of Ka indicates greater ionization thus more protons in solution, higher H+ concentration in solution means lower pH value.
Since HF have higher Ka have, thus HF solution will have lower pH.
The measurement is when the solution has the lower pH so HF should have higher ka.
Measurement when the solution has the lower pH?In the given solution, it contains the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq).
Here higher value of Ka represents the high ionization so there should be more protons in the solutions. and, the higher h+ concentration have the lower pH value.
So in the case when HF has higher Ka, due to this, HF solution will have lower pH.
Learn more about solution here: https://brainly.com/question/14619461
What causes the lines in the spectrum for elements
A tree can be a _________________________ for insects in a schoolyard.
Answer:
habitat/shelter
Explanation:
In the endothermic reaction below, how would adding heat affect the equilibrium of the system? (2 points)
A + B + energy ⇄ + D
a. shift it to the left toward the reactants
b. shift it to the right toward the products
c. equally create more products and reactants
d. it would have no effect
what is the difference between transmutation and natural radioactivity
Answer:
o diferença chave entre a radioatividade e a transmutação é que a radioatividade se refere à transmutação natural, enquanto a transmutação se refere à mudança de um elemento químico em outro por meios naturais ou artciais.Tanto a radioatividade quanto a transmutação são processos químicos que envolvem a mudança de núcleos atômicos para formar um novo elemento químico a partir de um elemento químico existente. A radioatividade é um tipo de processo de transmutação
Explanation:
mais informação em; ahttps://pt.strephonsays.com/radioactivity-and-transmutation-9500
I'm having with chemistry assignments I need a help
Answer:
the question please? illl help
In many cases, a negatively charged compound (ex Cl-) will interact with a positively charged metal (ex Fe2 ). Usually we would consider this type of interaction to be ionic, but in the case of metals we do not. What type of bond is being formed when a ligand interacts with a metal and how could you know that the bond you are forming is that type of bond
Answer:
See Explanation
Explanation:
An ionic bond occurs due to electrostatic attraction between a positively charged ion and a negatively charged ion.
A metal and a ligand are bound by a coordinate covalent bond or a dative bond. This bond occurs due to donation of electron pairs from ligands to available orbitals on metals.
The formation of coordinate bonds is evident when neutral molecules or negative ions with non bonding electrons donate same to empty metal orbitals. This is sometimes shown by an arrow pointing from the ligands to the metal center.
For instance; tetraammine copper II ion is formed when four ammonia molecules donate a lone pair each to available vacant orbitals of the copper metal center to form [Cu(NH3)4]^2+.
Did entropy increase or decrease?
1) 2 SO3 (g) → 2 SO2 (g)+O2 (g)
2) MgCO3 (s) + 2 H3O^+ (aq) → Mg^2+ (aq) + 3 H2 O(I) + CO2 (g)
Answer:
1) increase
2) increase
Explanation:
Entropy is the degree of disorderliness or randomness of a system. It is the measure of the unavailable energy in a system.
Entropy increases with increase in the number of particles. If the number of particles in a system increases from left to right, the entropy of the system increases accordingly.
In reaction 1, the number of particles from left to right increased from two to three hence the entropy was increased.
In reaction 2, the number of particles from left to right increased from three to five hence the entropy was increased.
Which property do all metals have?
А They are hard.
B They conduct electricity.
С They form acidic oxides.
D They react with water.
temperature usually increase when water condenses which behavior of water is most directly responsible for this phenomenon?
Answer:
Temperature usually increases when water condenses. What behavior of water is most directly responsible for this phenomenon? The release of heat by the formation of hydrogen bonds. Hydrogen bonds stabilize and keep the of ice farther apart than the water molecules of liquid water.
Liquid water - heat =
Pls help now!!!
Answer:
cold or ice?
Explanation:
have a good day.
i need help help me please
Answer:
C is the correct answer.
Explanation:
We can eliminate some answers immediately:
A is incorrect because nitrogen and hydrogen do not change their identity, they just combine into a new molecule that contains them. Notice how there are 2 nitrogen atoms and 6 hydrogen atoms on both sides of the equation.
B is incorrect because the first law of thermodynamics tells us that matter and energy will always be conserved, so the mass of the products must be equal to the mass of the reactants, not less than the mass of the reactants.
C is correct because balanced chemical equations help chemists predict how much product will form from certain amounts of reactants.
D is incorrect because this equation is showing the exact opposite. Molecular hydrogen and nitrogen can combine stoichiometrically to form ammonia.
20. Identi
21. Name the process and explain what is occurring at D.
Answer:
At D: Soil erosion us the process taking place.
Soil erosion is the washing away of the Earth's or soil's top layer by eroding agents.
1 Write the chemical symbols for following elements: Nitrogen-
Phosphorus- bromine- Zinc-Potassium-Magnesium.
Answer:
Look below
Explanation:
Nitrogen: N
Phosphorus: P
Bromine: Br
Zinc: Zn
Potassium: K
Magnesium: Mg
Hope this helps(would be very grateful if you mark this answer branliest bc i need for lvl up)
Arenediazoniums can undergo electrophilic aromatic substitutions with a wide variety of activated aromatic compounds to yield new azo dyes.
a. True
b. False
An alkaline earth hydroxide, M(OH)2, was taken to lab for analysis. The unknown powder was poured into a flask and swirled in room temperature DI water until a saturated solution formed. This solution was then slowly filtered to remove the undissolved solid hydroxide. 28.5 mL of this saturated solution was titrated with 0.173 M HCl (aq). Endpoint required 25.10 mL of the HCl (aq) solution. Calculate the Ksp for this alkaline earth hydroxide.
Answer:
1.77 * 10^-3
Explanation:
From the titration formula;
Let
CA = concentration of acid
CB = concentration of base
VA = volume of acid
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
The equation of the reaction is;
M(OH)2(aq) + 2HCl(aq) -------> MCl2(aq) + 2H2O(l)
So;
CAVA/CBVB= NA/NB
CAVANB = CBVBNA
CB= CAVANB/VBNA
CB= 0.173 * 25.10 * 1/28.5 * 2
CB= 4.3423/57
CB= 0.0762 M
This implies that the solubility of M(OH)2 = 0.0762 M
M(OH)2(s) ----> M^+(aq) + 2OH^-(aq)
So
Ksp = x * (2x)^2
Ksp = 4x^3
x = 0.0762
Ksp= 4(0.0762)^3
Ksp = 1.77 * 10^-3
C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion please show work
Answer:
[tex]\Delta H_{comb}=2043.85kJ/mol[/tex]
Explanation:
Hello there!
In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:
[tex]\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}[/tex]
Thus, given the values of the enthalpies of formation on the attached file, we obtain:[tex]-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol[/tex]
Best regards!