Answer:
i think is claim 1
Explanation:
moon jellies eat zooplankton and the zooplankton population increased
Answer:It’s important to know WHY. Great explanations are what makes Brainly awesomeExplain everything. Imagine that the person who was asking, knows nothing at all so you need to start from the very beginningAlways include the steps and/or background required to get to the final answer. Let’s help other people understand and solve future problems on their ownShow all the calculation if it is a numerical question.
Explanation:
what is an example of a change in genetic traits of an organism do to human affect
Answer:
A person's skin color, hair color, dimples, freckles, and blood type are all examples of genetic variations that can occur in a human population.
Explanation:
Estimate the volume of a solution of 5M NaOH that must be added to adjust the pH from 4 to 9 in 100 mL of a 100 mM solution of a phosphoric acid?
Answer:
3mL of 5M NaOH must be added to adjust the pH to 7.20
Explanation:
When NaOH is added to phosphoric acid, H₃PO₄, the reaction that occurs are:
NaOH + H₃PO₄ ⇄ NaH₂PO₄ + H₂O pKa1 = 2.15
NaOH + NaH₂PO₄ ⇄ Na₂HPO₄ + H₂O pKa2 = 7.20
NaOH + Na₂HPO₄ ⇄ Na₃PO₄ + H₂O pKa3 = 12.38
We can adjust the pH at 7.20 = pKa2 if NaH₂PO₄ = Na₂HPO₄. To make that, we must convert, as first, all H₃PO₄ to NaH₂PO₄ and the half of NaH₂PO₄ to Na₂HPO₄. To solve this question we need to find the moles of phophoric acid in the initial solution. 1.5 times these moles are the moles of NaOH that must be added to fix the pH to 7.20:
Moles H₃PO₄:
100mL = 0.100L * (0.100mol / L) = 0.0100 moles H₃PO₄
Moles NaOH:
0.0100 moles H₃PO₄ * 1.5 = 0.0150 moles NaOH
Volume NaOH:
0.0150 moles NaOH * (1L / 5moles) = 3x10⁻³L 5M NaOH are required =
3mL of 5M NaOH must be added to adjust the pH to 7.203 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
What is pH?It is the negative log of the concentration of Hydrogen ions in the solution.
To calculate the volume of NaOH first, calculate the moles of NaOH and H₃PO₄.
Moles of H₃PO₄.
[tex]\rm moles \ of \ H_3PO_4 = 100\rm \ mL = 0.100\rm \ L \times (0.100 \ mol / L)\\\\\rm moles \ of \ H_3PO_4 = 0.01[/tex]
The moles of NaOH:
[tex]\rm Moles \ of \ NaOH =0.01 \ moles \ H_3PO_4\times 1.5 \\\\\rm Moles \ of \ NaOH= 0.0150[/tex]
The volume of NaOH:
[tex]\rm Volume\ of \ NaOH = \rm 0.0150\ moles\ NaOH \times (1 \ L / 5 \ moles) \\\\\rm Volume\ of \ NaOH = 3\times 10^{-3} L[/tex]
Therefore, 3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.
Learn more about pH:
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Chemical bond stretching frequencies depend on two major factors, namely atomic weight and bond stiffness. The frequency absorbed is directly proportional to the bond strength (stiffness) but inversely proportional to the atomic weight. Which of the following bonds will give the highest absorption frequency value?
A. C single bond C.
B. C double bond C.
C. All would be the same frequency.
D. C triple bond C.
Answer:
D. C triple bond C.
Explanation:
The absorption frequency value is highest for carbon-carbon triple bond. The stretching frequencies are much higher than the bending frequency. It mainly depends upon the strength of the bonds as well as masses of the bonded atoms.
The triple bonds have higher stretching strength than the double bonds and double bonds have higher strength than single carbon to carbon bonds.
The triple bonds has a higher bond order, i.e. 3. So it has maximum stretching bonds.
chemical properties of citric acid
Answer:
Citric Acid is a weak acid with a chemical formula C6H8O7.
...
Properties of Citric Acid – C6H8O7.
C6H8O7 Citric Acid
Molecular Weight/ Molar Mass 192.124 g/mol
Density 1.66 g/cm³
Boiling Point 310 °C
Melting Point 153 °C
Hope it helps❤️
3.05 moles of H2O is equivalent to (blank amount) molecules of water.
Show your work.
Answer:
[tex]1 \: mole \: = 6.02 \times {10}^{23} \: molecules \\ 3.05 \: moles \: = (3.05 \times 6.02 \times {10}^{23} ) \: molecules \\ = 1.8361 \times {10}^{24} \: molecules[/tex]
At the start of a reaction, there are 0.0249 mol N2,
3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a
3.50 L reaction vessel at 375°C. If the equilibrium constant, K, for the reaction:
N2(g) + 3H2(g)= 2NH3(g)
is 1.2 at this temperature, decide whether the system is at equilibrium or not. If it is not, predict in which direction, the net reaction will proceed.
Answer:
Explanation:
The reaction is given as:
[tex]N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}[/tex]
The reaction quotient is:
[tex]Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
From the given information:
TO find each entity in the reaction quotient, we have:
[tex][NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}[/tex]
[tex][N_2] = \dfrac{0.024 }{3.5}[/tex]
[tex][N_2] = 0.006857[/tex]
[tex][H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}[/tex]
[tex][H_2] = 9.17 \times 10^{-3}[/tex]
∴
[tex]Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135[/tex]
However; given that:
[tex]K_c = 1.2[/tex]
By relating [tex]Q_c \ \ and \ \ K_c[/tex], we will realize that [tex]Q_c \ \ < \ \ K_c[/tex]
The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.
A 16.0 L sample contains 0.80 moles of a gas. If the gas is at the
same pressure and temperature, what is the number of moles when
the volume increases to 24.0 L? NOTE: You must show your
calculation on the attached scratch paper, including which of the
Gas Law formulas you used. *
A. 0.83 mol
B. 1.2 mol
C. 0.53 mol
D. 1.9 mol
(Please show your work)
[tex]\huge\bf\red{\underline{\underline{Answer:-}}}[/tex]
Here, 16 L = 0.80 Mole
Or, 1 L = ( 0.80 / 16 ) MoleOr, 1 L = 0.05 Mole_____________________________
Now, 24 L = 0.05 × 24 = [tex] \frac{5}{100} \times 24[/tex]= 120/100= 12/10= 1.2 Mole ( Ans )How many unpaired electrons are in the following complex ions?
a. Ru(NH3)62+ (low-spin case)
b. Ni(H2O)62+
c. V(en)33+
Solution :
A complex ion may be defined as a metal ion that is located at the center with other molecules or ions surrounding the center ion.
In the context, the number of the unpaired electrons for the following complex ions are :
1. [tex]$Ru(NH_3)62+$[/tex] (low spin)
[tex]$Ru^{2+} = [Kr] \ 4d^6 \ 5s^0$[/tex]
Here all the d-electrons are paired. Therefore, there are no unpaired electrons.
2. [tex]$Ni(H_2O)62+$[/tex] (high spin)
[tex]$Ni^{2+} = [Ar] \ 4s^0 \ 3d^8$[/tex]
Therefore, it has two unpaired electrons.
3. [tex]$V(en)33+$[/tex] (high spin)
[tex]$V=[Ar] \ 3d^2 \ 4s^0$[/tex]
It has 2 unpaired electrons
What is the function of a
catalyst?
A. Build enzymes
B. Speed up chemical reactions
C. Regulate the function of an enzyme
I believe it is to speed up chemical reactions
Which molecule will have double bonds?
(4 Points)
Fluorine
Nitrogen
Hydrogen
Oxygen
1. Calculate and interpret the equilibrium constant. Using the reaction below.
The equilibrium concentrations 0.60 M for E, 0.80 M for F, and 1.30 M for G. (Note: E, F, and G are all gases.) Do not include your solution.
Answer:
kc = [G]² / [E] [F], kc = [1.30M]² / [0.60M] [0.80M]
Explanation:
The reaction is:
E + F ⇄ 2G
The equilibrium constant, kc, must be written as the ratio of the molar concentrations of products over reactants. Each concentration powered to its coefficient.
For the reaction of the problem, kc is:
kc = [G]² / [E] [F]Replacing the given concentrations:
kc = [1.30M]² / [0.60M] [0.80M]What is the pH of a 6.2 x 10-5 M NaOH solution?
Answer:
4.21Explanation:
The pH of a solution can be found by using the formula
[tex]pH = - log [ {H}^{+} ][/tex]
From the question we have
[tex]ph = - log(6.2 \times {10}^{ - 5} ) \\ = 4.207608[/tex]
We have the final answer as
4.21Hope this helps you
If the plant population decreased, the amount of carbon in the atmosphere would _______.
Increase
Stay the same
Decrease
Answer:
increase answer
Explanation:
i hope that is right
A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pressure of the gas after it is heated to 278 K?
Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.
Explanation:
Given: [tex]T_{1}[/tex] = 178 K, [tex]P_{1}[/tex] = 0.3 atm
[tex]T_{2}[/tex] = 278 K, [tex]P_{2}[/tex] = ?
According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.
Formula used to calculate the pressure is as follows.
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\[/tex]
Substitute the values into above formula is as follows.
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm[/tex]
Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.
A 0.750 M sample of phosphorus pentachloride decomposes to give phosphorus trichloride and chlorine gases.If the equilibrium concentration of PCl5 is 0.650M,what is the equilibrium constant for the reaction?
PCl5(g) --------> PCl3(g)+ Cl₂(g)
A) Keq = 0.0133
B) Keq = 0.0154
C) Keq = 0.133
D) Keq = 0.154
E) Keq = 65.0
Answer:
Explanation:
Amount of phosphorus pentachloride reacted = 0.75 M - 0.65 M = 0.10 M
PCl₅(g) --------> PCl₃(g)+ Cl₂(g)
1 mole 1 mole 1 mole
Amount of phosphorus pentachloride reacted = 0.75 M - 0.65 M = 0.10 M
PCl₃(g) and Cl₂(g) formed are .10 M and .10 M
Keq = [ PCl₃] x [ Cl₂ ] / [ PCl₅ ]
= .10 M x .10 M / .65 M
= .0154
B ) is the correct choice .
Enthalpy change is the
a. amount of energy absorbed or lost by a system as heat during a process at constant pressure.
b. entropy change of a system at constant pressure.
c. pressure change of a system at constant temperature.
d. temperature change of a system at constant pressure.
Answer:
I believe is A
Explanation:
Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure.
Thermal energy naturally flows from _________ matter to _______ matter.
Answer:
Warmer
Cooler
Explanation:
What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?
Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
Explanation:
Given: [tex]V_{1}[/tex] = 61 L, [tex]T_{1}[/tex] = 183 K, [tex]P_{1}[/tex] = 0.60 atm
At STP, the value of pressure is 1 atm and temperature is 273.15 K.
Now, formula used to calculate the new volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L[/tex]
Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
when 18.0 g H20 is mixed with 33.5 g Fe, which is the limiting reactant?
Answer:
si.mple fe
Explanation:
ghhsshzhzbbzhh
How many grams of carbon are In 25 g of CO2?
Answer: almost 7g
Explanation: 25 grams of carbon dioxide contains 6.818 grams of carbon.
Which of the following statement is true?
a. The resonance effect of the hydroxyl group stabiizes the anionic intermediate.
b. The resonance effect of the hydroxyl group stabilizes the cationic
c. The inductive effect of the hydroxyl group stabiizes the cationic intermediate
d. The inductive effect of the hydroxyl group stabiizes the anionic intermediate
Answer:
a. The resonance effect of the hydroxyl group stabilizes the anionic intermediate
Explanation:
The resonance effect stabilizes the the charge through the delocalization of the pi bonds. The resonance stabilization mainly occurs in the conjugated pi systems.
For example, phenol forms a strong hydrogen bonds than the nonaromatic alcohols as the [tex]$O_2-H_2$[/tex] dipole present in the hydroxyl group is being stabilized by the presence of the aromatic ring of phenol.
Thus the resonance effect of the hydroxyl group stabilizes the anionic intermediate.
A container holds 40.0 mL of nitrogen at 30° C and at a constant pressure.
Find its volume if the temperature increases to 80° C?
Answer:
The correct answer is - 46.60 mL.
Explanation:
To find the volume of the gas at its new increased temperature we need to use Charl Law that shows the direct relationship between Volume and Temperature while Pressure remains constant.
V1 = 40 ml
T1 = 30 degree C + 273 = 303 K
V2 = ?
T2 = 80 degree C + 273 = 353 K
Charl Equation is:
V 1/T 1 = V 2/ T 2
(V1) * (T2)/ T1= V2
placing value:
40*353/303 = V2
= 14120/303
Vf = 46.60 mL
n-hexane boils at 68.7 oC at 1.013 bar and the heat of Vaporization is 28850
J/mol. Calculate Its Entropy ?
how to solve x² in differential
Answer:
x² = mutiphy by them self
Explanation:
The blending together of some genes is called:
Answer:
its called molding
Explanation:
Calculate the energy of an electron in the n = 2 level of a hydrogen atom.
Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
Explanation:
Given: n = 2
The relation between energy and [tex]n^{th}[/tex] orbit of an atom is as follows.
[tex]E = - \frac{13.6}{n^{2}} eV[/tex]
Substitute the values into above formula as follows.
[tex]E = - \frac{13.6}{n^{2}} eV\\= - \frac{13.6}{(2)^{2}}\\= - 3.40 eV[/tex]
The negative sign indicates that energy is being released.
Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
0=4
Balance this equation
H₂sicl2+ H₂O → H8Si4O4 + HCl
Balanced Equation is
4H2SiCl2+4H2O → H8Si4O4 + 8HCl
Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen,
and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50
Answer:
C₃H₈O
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.031 g
Mass of CO₂ = 2.265 g
Mass of H₂O = 1.236 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:
For carbon, C:
Mass of CO₂ = 2.265 g
Molar mass of CO₂ = 44.01 g/mol
Molar mass of C = 12.01 g/mol
Mass of C =?
Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂
Mass of C = 12.01/44.01 × 2.265
Mass of C = 0.618 g
For hydrogen, H:
Mass of H₂O = 1.236 g
Molar mass of H₂O = 18.02 g/mol
Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol
Mass of H =?
Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O
Mass of H = 2.02/18.02 × 1.236
Mass of H = 0.139 g
For oxygen, O:
Mass of compound = 1.031 g
Mass of C = 0.618 g
Mass of H = 0.139 g
Mass of O =?
Mass of O = Mass of compound – (mass of C + mass of H)
Mass of O = 1.031 – ( 0.618 + 0.139)
Mass of O = 1.031 – 0.757
Mass of O = 0.274 g
Finally, we shall determine the empirical formula. This can be obtained as follow:
C = 0.618 g
H = 0.139 g
O = 0.274 g
Divide by their molar mass
C = 0.618 / 12.01 = 0.051
H = 0.139 / 1.01 = 0.138
O = 0.274 / 16 = 0.017
Divide by the smallest
C = 0.051 / 0.017 = 3
H = 0.138 / 0.017 = 8
O = 0.017 / 0.017 = 1
Therefore, the empirical formula of the compound is C₃H₈O
Based on the data provided, the empirical formula of the compound is C₃H₈O
What is empirical formula?The empirical formula of compound is its simplest formula showing the mole ratio of the elements in the compound.
First, the mass of carbon, hydrogen and oxygen in the compound is determined first as follows:
For carbon, C:
Mass of CO₂ = 2.265 g
Molar mass of CO₂ = 44.01 g/mol
Molar mass of C = 12.01 g/mol
Mass of C = 12.01/44.01 × 2.265
Mass of C = 0.618 g
For hydrogen, H:
Mass of H₂O = 1.236 g
Molar mass of H₂O = 18.02 g/mol
Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol
Mass of H = 2.02/18.02 × 1.236
Mass of H = 0.139 g
For oxygen, O:
Mass of compound = 1.031 g
Mass of C = 0.618 g
Mass of H = 0.139 g
Mass of O = Mass of compound – (mass of C + mass of H)
Mass of O = 1.031 – ( 0.618 + 0.139)
Mass of O = 0.274 g
To determine the empirical formula of a compound, the mole ratio of the elements are determined as follows:
Mole ratio = reacting mass/molar massMole ratio of the elements:
Carbon Hydrogen Oxygen
0.618 g/12.01 0.139 / 1.01 0.274 / 16
0.051 0.138 0.017
Divide by the smallest ratio to convert to whole numbers
0.051 / 0.017 0.138 / 0.017 0.017 / 0.017
3 : 8 : 1
Therefore, the empirical formula of the compound is C₃H₈O
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What is the Ksp expression for the dissociation of calcium oxalate?Immersive Reader
(4 Points)
Ksp=[Ca⁺²] x [C₂O₄⁻²]
Ksp=[Ca⁺²]² x [C₂O₄⁻²]
Ksp=[Ca⁺²]⁴ x [C₂O₄⁻²]
Ksp=[Ca⁺²] x [C₂O₄⁻²]²
Answer:
Ksp = [Ca⁺²] × [C₂O₄⁻²]
Explanation:
Step 1: Write the balanced reaction for the dissociation of calcium oxalate
CaC₂O₄(s) ⇄ Ca⁺²(aq) + C₂O₄⁻²(aq)
Step 2: Write the expression for the solubility product constant (Ksp) of calcium oxalate
The solubility product constant is the equilibrium constant for the dissociation reaction, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It doesn't include solids nor pure liquids because their activities are 1.
Ksp = [Ca⁺²] × [C₂O₄⁻²]
Which of the following
describes the zone of the
ocean where no light reaches?
A. up to 200 meter depth and includes
photosynthetic plants, sea anemones,
sponges, crabs, and clams
B. the "twilight zone" between 200-1000
meters deep and includes whales and octopi
and little life
C. permanent darkness below 1000 meters
with bioluminescent bacteria, bottom
feeders, and angler fish
Answer:
Bathypelagic
54% of the ocean lies in the Bathypelagic (aphotic) zone into which no light penetrates. This is also called the midnight zone and the deep ocean. Due to the complete lack of sunlight, photosynthesis cannot occur and the only light source is bioluminescence.
Explanation:
The small surface zone that has light is the photic zone. The entire rest of the ocean does not have light and is the aphotic zone.
Permanent darkness below 1000 meters with bioluminescent bacteria, bottom feeders, and angler fish is where no light reaches.
What is Darkness?This is referred to the state of being dark as a result of absence of light in the area.
The light ray penetration decreases with increase in depth thereby making areas below 1000 meters dark with bioluminescent bacteria, bottom feeders, and angler fish which is why option C was chosen.
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