Answer: The amino acid that has a positively charged R group are LYSINE and ARGININE.
Explanation:
AMINO ACIDS are the basic structural units of proteins. Each amino acid contains an amino group ( -NH2) and a carboxyl group (-COOH) in its molecule. The carbon atom of amino acid to which the functional groups are attached is know as the alpha-carbon. In neutral solution, amino acids are mainly in the form of dipolar ions. Amino acids can be prepared through hydrolysis of protein by boiling with dilute hydrochloric acid.
Amino acids can be classified according to the chemical properties of its side chain (R group) at pH 7, these include:
--> Amino acids with POSITIVELY CHARGED R group
--> Amino acids with negatively charged R group
--> Amino acids with neutral polar R group
--> Amino acids with nonpolar aliphatic R group.
Amino acids with POSITIVELY CHARGED R group are those amino acids that has side chains which contain nitrogen and resemble ammonia, which is a base. Their pKa's are high enough that they tend to bind protons, gaining a positive charge in the process. Example include LYSINE and ARGININE.
How many milliliters of a 6.00 M NaCl solution are needed to make 250.0 milliliters of a 0.500 M NaCl solution?
Answer:
20,8ml NaCl 6M
Explanation:
C1V1 = C2V2
Which of the following colors has the highest energy? O A. Red O B. Green O C. Blue O D. Yellow
Answer:
C. BlueExplanation:
This is because, Blue color highest frequency of energy after Violet and Indigo.
What are the two types of addition compounds
Answer:
electrophilic addition and
nucleophilic addition.
Answer:
the two types of addition compoundsare:
1.electrophilic addition
2. nucleophilic addition.
Sợ sánh kỹ thuật trích ly rắn lỏng với kỹ thuật trích ly lỏng
lỏng
Answer:
which language is this please tell me
Classify each structure according to its functional class.
Compound A contains a carbonyl bonded to two alkyl groups.
Compound B contains an oxygen bonded to two alkyl groups.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound D is a nitrogen bonded to three alkyl groups.
Classify structure A according to its functional class.
Classify structure B according to its functional class.
Classify structure C according to its functional class.
Classify structure D according to its functional class.
Answer:
Classify each structure according to its functional class.
Compound A contains a carbonyl bonded to two alkyl groups.
Compound B contains an oxygen bonded to two alkyl groups.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound D is a nitrogen bonded to three alkyl groups.
Explanation:
Compound A contains a carbonyl bonded to two alkyl groups.
-C=O group is called a carbonyl group.
If it is present between two alkyl groups then, it is a ketone.
Compound B contains oxygen bonded to two alkyl groups.
Compound B is an example of an ether molecule.
Compound C contains a carbonyl bonded to propyl and N H C H 3.
Compound C is C3H7-CO-NHCH3 which is an amide molecule.
Compound D is nitrogen bonded to three alkyl groups.
This is an example of a tertiary amine group.
____________ can increase the presence of 5HT in the terminal button or synaptic cleft.
Answer:
Selective serotonin reuptake inhibitors (SSRIs)
Explanation:
A synaptic cleft is a space that separates two neurons thereby forming a junction between two or more neurons. The synaptic cleft helps in the transfer of nerve impulse from one neuron to the other.
5-HT is found in the enteric nervous system located in the gastrointestinal tract and it helps in modulating cognition, memory, sleep, and numerous physiological processes.
Selective serotonin reuptake inhibitors (SSRIs), such as fluoxetine and citalopram are used to increase the level of 5-HT in the synaptic cleft by inhibiting its reuptake into the presynaptic terminal.
Select the net ionic equation for the reaction that occurs when magnesium sulfate and nickel(II) nitrate are mixed.
a. Ni2+(aq) + SO4^2- → NISO2 (s) + O2 (g).
b. Mg2+(aq) + 2NO3 (aq) → Mg(NO3)2(s).
c. Mg2+(aq) + NO3- → MgNO3 (s).
d. Mg2+(aq) + SO4^2- (aq) + Ni2+ (aq) + 2NO3- → Mg2+ (aq) + 2NO3 (aq) + NISO4 (s).
e. Ni2+(aq) + SO4^2- (aq) → NISO4 (s).
f. No reaction occurs.
Answer:
No reaction occurs.
Explanation:
The molecular reaction is as follows;
MgSO4(aq) + Ni(NO3)2(aq) ----> Mg(NO3)2(aq) + NiSO4(aq)
We can see from the reaction above that the both products of the reaction are soluble. Recall that a double replacement reaction often yields one insoluble product which separates as a precipitate.
This reaction does not occur since the two products that ought to be obtained are soluble in water.
The combustion of ethylene proceeds by the reaction
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (g)
When the rate of disappearance of C2H4 is 0.13 M s-1, the rate of appearance of CO2 is ________ M s-1.
What is the quantity of
heat required to raise the
temperature of 500 g of
iron by 2°C?
The specific heat capacity
of iron is 500 J/(kg °C)
Answer:
The quantity of heat required to raise the temperature of 500 g of iron by 2°C is 500 J.
Explanation:
Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state (solid, liquid or gaseous).
Its mathematical expression is the fundamental equation of calorimetry. This is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q= ?c= 500 [tex]\frac{J}{kg*C}[/tex]m= 500 g= 0.500 kgΔT= 2 CReplacing:
Q= 500 [tex]\frac{J}{kg*C}[/tex] *0.500 kg*2 C
Solving:
Q= 500 J
The quantity of heat required to raise the temperature of 500 g of iron by 2°C is 500 J.
mẫu khi thêm NH4NO3 vào đem nung để nguội lại thêm NH4NO3 có tác dụng gì?
Answer:
Adding ammonium nitrate to water turns the mixture cold and is a good example of an endothermic chemical reaction!
Chromosomes contain the genetic material found in cells. true or false?
1. true
2. false
Answer:
1) True
Explanation:
.What shape does each galaxy have?
Answer:
as in shapes
Explanation:
The reaction below was carried out in an acidic solution l-+10-3--->l2 Which statement is true about this equation? Although it is unbalanced, it can be balanced by using the half-reaction method It has been balanced by using the half-reaction method. It has been balanced by directly using spectator ions. O Although it is unbalanced, it can be balanced by directly using spectator ions.
Answer:
answer is b. : It has been balanced by using the half-reaction method.
Explanation:
edge 2021
Answer:b
Explanation:cuz it is
The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) list carbon only Chirulit,
Answer:
(d)
Explanation:
Carbonyl group can be the placement of kerosene sugar
65. Which of the following heavy metal is not
toxic?
(1 Point)
Hg
Pb
O As
O Ni
Which event is an example of melting?
A. Wax drips down the side of a lit candle.
B. Perspiration dries on a person's skin.
C. Rain turns to ice pellets.
D. A mirror fogs up when someone takes a hot shower.
I’m just curious tbh
Answer:
A. Wax drips down the side of a lot candle.
Explanation:
The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them
Answer:
A. Wax drips down the side of a lot candle.
Explanation:
please answer all if you can!
Answer:1.b 2.A (i think it might be c) 3 (i have no clue) 4.(no clue) Sorry i only got 1 and 2 witch i think are right its been a while sense i have done this....
Explanation:
Write a balanced equation for the complete combustion of 2,3,3-trimethylpentane. Use the molecular formula for the alkane (C before H) and the smallest possible integer coefficients.
Answer:
C8H18 + 25/2O2 ----> 8CO2 + 9H2O
Explanation:
2,3,3-trimethylpentane has the molecular formula C8H18.
The general formula for the combustion of an alkane is;
CnH2n+2 + 3n+1/2O2 ----->nCO2 + (n+1)H2O
In writing a balanced chemical reaction equation, the number of atoms of each element on the left hand side of the reaction equation must be the same as the number of atoms of the same element on the right hand side of the reaction equation.
For C8H18, the balanced chemical reaction equation for combustion is;
C8H18 + 25/2O2 ----> 8CO2 + 9H2O
Identify the conjugate acid/base pairs in each of the following equations:
(a) H2S + NH3 ⇔ NH4+ + HS-
Pair 1: H2S and
Pair 2: NH3 and
(b) HSO4- + NH3 ⇔ SO42- + NH4+
Pair 1: HSO4- and
Pair 2: NH3 and
(c) HBr + CH3O- ⇔ Br- + CH3OH
Pair 1: HBr and
Pair 2: CH3O- and
(d) HNO3 + H2O → NO3- + H3O+
Pair 1: HNO3 and
Pair 2: H2O and
Answer:
(a) Pair 1: H₂S and HS⁻
Pair 2: NH₃ and NH₄⁺
(b) Pair 1: HSO₄⁻ and SO₄⁻
Pair 2: NH₃ and NH₄⁺
(c) Pair 1: HBr and Br⁻
Pair 2: CH₃O⁻ and CH₃OH
(d) Pair 1: HNO₃ and NO₃⁻
Pair 2: H₃O⁺
Explanation:
When an acid loses its proton (H⁺), a conjugate base is produced.
When a base accepts a proton (H⁺), it forms a conjugate acid.
(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.
NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺
(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.
The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.
(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.
CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.
(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.
H₂O gains a proton to form the conjugate acid H₃O⁺.
Identify the item that does not have kinetic energy.
A. A worm crawling
B. A falling leaf
C. An airplane flying
D. A butterfly sitting on a twig
Answer:
D
Explanation:
there is no other answer choice that does not move. This is a fool-proof question because all the other answer choices contain movement except the butterfly resting
why Mg(OH)2 is soluble in HCL
Answer:
While Mg(OH)2 is practically insoluble, a certain amount of Mg(OH)2 dissociates into ions when put in water. ... As HCl is added to the beaker containing milk of magnesia, the H+ ions from the HCl react with the OH– ions (those that are actually in solution from the Mg(OH)2) according to Equation 3 below.
Fe có tác dụng với HCL không
Fe có tác dụng với Hcl
Fe + 2hcl -> fecl2 +h2
Answer:
có
Explanatio:
He predicted an element with an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum that he named ekaboron. This element was discovered in 1879. What is this element:?
Answer:
Scandium
Explanation:
Mendeleev played an important role in the development of the modern periodic table. His periodic table was filled with gaps. He said that these gaps were elements that were yet to be discovered. He rightly predicted many elements which have now been discovered and fitted in their proper places in the periodic table.
He used the prefix ''eka'' to refer to elements whose properties were alike but were yet to be discovered at that time.
The compound named ekaboron which he predicted to have an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum was later discovered in 1879 and properly named scandium.
Which best describes how the total mass of the substances that go into
photosynthesis compares to the mass of substances that are present
afterward?
O A. The mass increases because the molecules that are produced are
larger than those that are used.
B. The mass increases because some light energy changes into
mass.
O C. The mass stays the same because the total number of atoms
does not change
O D. The mass decreases because plants destroy some of the atoms
during photosynthesis.
Answer:
C. The mass stays the same because the total number of atoms does not change
Explanation:
According to the law of conservation of matter/mass, matter cannot be created nor destroyed, hence, the amount of matter in the reactants must be the same amount in the products.
Using the photosynthetic reaction as a case study, carbon dioxide (CO2) and water (H2O) are the compounds that go into the reaction (reactants) while glucose and oxygen (O2) are the products of the reaction.
Using the law of conservation of matter to explain, the total mass of both the reactants and products stays the same because the total number of atoms does not change i.e. if 6 atoms of Carbon starts the reaction, 6 atoms of carbon will end it.
A sample of drinking water was tested for Pb2 and was found to have a Pb2 concentration of 17.50 ppb. A 9.00 mL sample of the drinking water was spiked with 1.00 mL of a 2.29 ppb Pb2 standard. Analysis of the spiked sample gave a concentration of 15.93 ppb Pb2 . Find the percent recovery of the spike.
Answer:
[tex]X=75\%[/tex]
Explanation:
From the question we are told that:
Concentration [tex]C_1=17.50ppb[/tex]
Volume [tex]v=9mL[/tex]
Spike Volume [tex]V_s=1.00mL[/tex]
spike Conc [tex]C_2=2.29 ppb[/tex]
Analysis Conc [tex]C_s=15.93 ppb[/tex]
Generally the equation for percent recovery is mathematically given by
[tex]X=\frac{C_s'-C_1'}{C_2'}[/tex]
Where
Concentration of spiked sample C_s'
[tex]C_s'=C_s*(v+v_s)[/tex]
[tex]C_s'=15.92*(9+1)[/tex]
[tex]C_2=159.2[/tex]
Concentration of unspiked sample C_1'
[tex]C_1'=17.50*9mL[/tex]
[tex]C_1'=157.5[/tex]
Concentration of spike sample C_2'
[tex]C_2=2.29*1[/tex]
[tex]C_2=2.29[/tex]
Therefore
[tex]X=\frac{159.2-157.5}{2.29}[/tex]
[tex]X=75\%[/tex]
For a particular chemical reaction the rate (g/hr) at which one of the reactants changes is proportional to the amount of that reactant present. If y represents the amount of that reactant at time t, StartFraction dy Over dt EndFraction equals minus0.7y. If there were 70 grams of the reactant when the process started (tequals 0), how many grams will remain after 4 hours?
Answer:
Amount of reactant after four hours = 4,26 grams
Explanation:
Suppose y denotes the amount of reactant at the time (t)
The given function:
[tex]\dfrac{dy}{dt} = -0.7 y[/tex]
[tex]\dfrac{dy}{y} = -0.7 dt[/tex]
Taking integral on both sides
㏑(y) = -0.7t + c¹
[tex]e^{In(y)}= e^{-0.7t + c^1}[/tex]
[tex]y(t) = Ce ^{-0.7t}[/tex]
At t = 0 ; y (t) = 70
∴
[tex]70 = Ce^{-0.7(0)}[/tex]
C = 70
As such; [tex]\mathtt{y(t) = 70 e^{-0.7*t}}[/tex]
After four hours, the amount of the reactant is:
[tex]\mathtt{y(t) = 70 e^{-0.7*4}}[/tex]
[tex]\mathtt{y(t) = 70 e^{-2.8}}[/tex]
[tex]\mathtt{y(t) = 4.26}[/tex]
Amount of reactant after four hours = 4,26 grams
PLZ HELP!!!! NO LINKS
Answer:
the answer would be 6.720
Simply subtract the combined mass by the mass of the beaker.
The answer has 3 decimal places because the least amount of decimal places in the equation is 3.
please help! what is the correct answer to this picture
Answer:
i think its c [everything is so blury]
Explanation:
Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.
Answer:
14.9 g
Explanation:
Step 1: Write the balanced equation
C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O
Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀
The molar mass of C₄H₁₀ is 58.12 g/mol.
9.57 g × 1 mol/58.12 g = 0.165 mol
Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀
0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O
Step 4: Calculate the mass corresponding to 0.825 mol of H₂O
The molar mass of H₂O is 18.02 g/mol.
0.825 mol × 18.02 g/mol = 14.9 g
A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the water? (Specific heat capacity of water = 4.184 J/goC.)
Answer: The final temperature will be [tex]52.74^oC[/tex]
Explanation:
Calculating the heat released or absorbed for the process:
[tex]q=m\times C\times (T_2-T_1)[/tex]
In a system, the total amount of heat released is equal to the total amount of heat absorbed.
[tex]q_1=-q_2[/tex]
OR
[tex]m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2)[/tex] ......(1)
where,
C = heat capacity of water = [tex]4.184J/g^oC[/tex]
[tex]m_1[/tex] = mass of water of sample 1 = 100.0 g
[tex]m_2[/tex] = mass of water of sample 2 = 71.0 g
[tex]T_f[/tex] = final temperature of the system = ?
[tex]T_1[/tex] = initial temperature of water of sample 1 = [tex]27^oC[/tex]
[tex]T_2[/tex] = initial temperature of the water of sample 2 = [tex]89.0^oC[/tex]
Putting values in equation 1, we get:
[tex]100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC[/tex]
Hence, the final temperature will be [tex]52.74^oC[/tex]