The structure of amylase deteriorates due to high temperature of the solution.
This experiment shows that the structure of amylase deteriorates due to high temperature which prevents this amylase from performing its function properly.
At high temperatures the amylase will break starch down very slowly or not at all due to denaturation of the enzyme's active site due to which it can't perform its function properly so we can conclude that high temperature denatures amylase enzyme.
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Nakula's conclusion was- "My results show that boiling destroys amylase"
Amylase is an enzyme that breaks down starch molecules into sugar molecules. He boiled the solution in tube P, and when he checked tube P for sugar, there wasn't any. He didn't boil the solution in tube Q and he found sugar in it.
Amylase had broken down starch molecules to sugar molecules in tube Q. Tube P's solution had been boiled, and this showed that when he boiled it, it destroyed the amylase, that is why the starch molecules hadn't been broken down into sugar molecules.
A local am radio station broadcasts at a wavelength of 349m, calculate the energy of the wavelength at which it is broadcasting?a. 5.69 x 10^-19 J b. 6.93 x 10^-39 J c. 8.59 x 10^5 J d. 6.33 x10^-45 J e. 7.71 x 10^-40 J
Answer:
E = 5.69x10⁻²⁸m
Explanation:
To solve this question we neeed to convert the wavelength in meters to energy in joules using the equation:
E = hc / λ
Where E is energy in joules, h is Planck's constant = 6.626x10⁻³⁴Js
c is light constant = 3.0x10⁸m/s
And λ is wavelength in meters = 349m
Replacing:
E = 6.626x10⁻³⁴Js*3.0x10⁸m/s / 349m
E = 5.69x10⁻²⁸m
The energy of the wavelength at which the local AM radio station is broadcasting is: A. [tex]5.69 \times 10^{-19}\;J[/tex]
Given the following data:
Wavelength = 349 nanometer = [tex]3.49 \times 10^{-7}\;meter[/tex]Scientific data:
Speed of light = [tex]3 \times 10^8\;meters[/tex]Planck constant = [tex]6.626 \times 10^{-34}\;J.s[/tex]To calculate the energy of the wavelength at which the local AM radio station is broadcasting, we would apply Einstein's equation for photon energy:
Mathematically, Einstein's equation for photon energy is given by the formula:
[tex]E=\frac{hv}{\lambda}[/tex]
Where:
E is the energy.h is Planck constant. [tex]\lambda[/tex] is the wavelength.v is the speed of light.Substituting the given parameters into the formula, we have;
[tex]E=\frac{6.626 \times 10^{-34}\times 3 \times 10^8}{3.49 \times 10^{-7}}\\\\E=\frac{1.99 \times 10^{-25}}{3.49 \times 10^{-7}} \\\\E=5.69 \times 10^{-19}\;Joules[/tex]
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How many colors are there in a rainbow?
[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]
There are 7 colours in a rainbow The colours of the rainbow are Red, Orange, Yellow, Green, Blue, Indigo and Violet.Explanation:
there r seven colors in a rainbow.red, orange, yellow, green, blue, indigo and violet.hope it helps.stay safe healthy and happy..A Chef fills out a 50mL container with 43.5g of cooking oil, What is the density of the oil?
Answer:
.87
Explanation:
p = m/V
43.5/50
.87
sự sắp xếp nguyên tử trong vật chất
Answer:
sosksjsjjs
Explanation:
even i know how to type şïllily
Monomers that each contain a 5-carbon sugar, a phosphate group, and a nitrogenous base combine and form which type of polymer?
A. Amino acid
B. Carboxylic acid
C. Nucleic acid
D. Fatty acid
Answer:
The correct answer is C. Nucleic acid
Explanation:
Nucleic acids are biological polymers which play an important role in the storage and expresion of genetic information. There are two types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are basically composed of:
- a 5-carbon sugar: deoxyribose in DNA and ribose in RNA
- phosphate group
- a nitrogenous base: adenine, cytosine, guanine and thymine in DNA; while RNA contains adenine, cytosine, guanine and uracil.
Which system monitors carbon dioxide levels in
the blood?
There are two kinds of respiratory chemoreceptors: arterial chemoreceptors, which monitor and respond to changes in the partial pressure of oxygen and carbon dioxide in the arterial blood, and central chemoreceptors in the brain. hope this helps you
have a nice day
Ethanol is the alcohol found in brandy, that is sometimes burned over cherries to make the dessert cherries jubilee. Write a balanced equation for the complete oxidation reaction that occurs when ethanol (C2H5OH) burns in air. Use the smallest possible integer coefficients.
Answer: The balanced equation for the complete oxidation reaction of ethanol is [tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]
Explanation:
Combustion is the chemical process where an organic molecule reacts with oxygen gas present in the air to produce carbon dioxide and water molecules.
It is also known as an oxidation reaction because oxygen is getting added.
The chemical equation for the oxidation of ethanol follows:
[tex]C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O[/tex]
By stoichiometry of the reaction:
1 mole of ethanol reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide gas and 3 moles of water molecules.
2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.
Answer:
Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).
Explanation:
Hello there!
In this case, according to the given redox reaction, we rewrite it as a convenient first step:
[tex]2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O[/tex]
Next, we assign the oxidation numbers as follows:
[tex]2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}[/tex]
Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).
Regards!
Question 9 of 25
How many hydrogen atoms are in a molecule of table sugar (C12H,2011)?
O A. 12
B. 45
C. 11
D. 22
SUBMIT
D.22
is my answer than welcome
The following properties are either physical or chemical. Which one is different from the rest based on those two categories? (5 points)
Boiling point
Density
Ductility
4)
Answer:
The following properties are either physical or chemical. Which one is different from the rest based on those two categories? We chose all of the above
What was the plum pudding atomic model
What is chalk considered?
Explanation:
Chalk, soft, fine-grained, easily pulverized, white-to-grayish variety of limestone. Chalk is composed of the shells of such minute marine organisms as foraminifera, coccoliths, and rhabdoliths. The purest varieties contain up to 99 percent calcium carbonate in the form of the mineral calcite.
Answer:
Chalk is a salt because it has a calcium base; you can tell that it is a salt because it is powdery. It's formula is CaCO₃, or calcium carbonate.
A 0.334 g sample of an unknown compound occupies 245 ml at 298 K and 1.22 atm. What is the molar mass of the unknown compound.
Answer:
27.4 g/mol
Explanation:
Assuming the compound is a gas and that it behaves ideally, we can solve this problem by using the PV=nRT formula, where:
P = 1.22 atmV = 245 mL ⇒ 245 mL / 1000 = 0.245 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 298 KInputting the data:
1.22 atm * 0.245 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 KSolving for n:
n = 0.0122 molWith the calculated number of moles and given mass, we calculate the molar mass:
0.334 g / 0.0122 mol = 27.4 g/molIn the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by
A) NaF
B) MgF₂
C) MgBr₂
D) AlF₃
E) AlBr₃
In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct
What factors affect the magnitude of energy of ionic crystalline solids ?For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.
Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase
Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.
The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids
Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct
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energy difference between t2g and EG level in octahedral complex is denoted by
Answer:
∆o
Explanation:
Usually, the d level consists of a set of five degenerate d orbitals. These orbitals rema degenerate until the approach of ligands.
When ligands approach, the d orbitals are split into higher energy eg and lower energy t2g orbitals.
There exists an energy difference between this t2g and eg levels. This energy difference is called the octahedral crystal field splitting denoted as ∆o.
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
Convert 0.2502 grams of calcium nitrate to moles.
How many atoms are contained in 3.80 moles of magnesium?
Answer:
a) 0.001525 mol Ca(NO₃)₂
b) 2.29 × 10²⁴ atoms Mg
General Formulas and Concepts:
Atomic Structure
Reading a Periodic Table Reactions rxn CompoundsExplanation:
a)
Calcium (Ca) has a +2 charge on the periodic table. Polyatomic ion for nitrate (NO₃) has a -1 charge. We need to balance the compound:
Ca(NO₃)₂
To find the molar mass of calcium nitrate, we must pull data from the PT.
Ca has a molar mass of 40.08 g/mol
N has a molar mass of 14.01 g/mol
O has a molar mass of 16.00 g/mol
Add these all together and multiply respectively to find the molar mass of the compound:
40.08 + 2[14.01 + 3(16.00)] = 164.1 g/mol
To find the amount of moles we have given grams, divide the grams by the molar mass:
(0.2502 g) / (164.1 g/mol) = 0.001525 mol Ca(NO₃)₂
We are given 4 sig figs and our final answer is in 4 sig figs.
b)
We need to know Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc. This converts moles into atoms, molecules, formula units, etc.
We are given 3.80 moles of Mg. To find the number of atoms, we multiply Avogadro's number:
3.80 mol Mg · (6.022 × 10²³ atoms Mg/ 1 mol Mg) = 2.28836 × 10²⁴ atoms Mg
Since we are given 3 sig figs, we must round:
2.28836 × 10²⁴ atoms Mg ≈ 2.29 × 10²⁴ atoms Mg
Answer:
Solution given:
a.
we have
1 mole of calcium nitrate=164.088g
now
0.2502g of calcium nitrate=1/164.088*0.2502=
0.0015248 mole of calcium nitrate
0.2502 grams of calcium nitrate
=0.0015248moles.
b.
1 mole of magnesium=[tex]6.02214076 × 10^{23}[/tex]atoms
3.80 mole of magnesium=[tex]6.02214076 × 10^{23}×3.80=2.2884*10^{24}[/tex]atoms
[tex]\bold{\underline{2.2884*10^{24}}}[/tex]atoms
are contained in 3.80 moles of magnesium.
1. Choose binary compounds with ionic bonds:
a) CCl4; b) KCl; c) ZnO; d) SiO
Answer:
its kcl ,option.b)
Explanation:
because its contains k+and cl- which is and ionic bond
URGENT- please do by 14th July if possible!!!
1. How do metals react with acids?
2. What are the similarities and differences in the way different metals react with water and acids?
3. Why are some metal is more reactive than others
4. Why is the reactivity of metals so important to us?
5. What the displacement reactions?
6. Why do you displacement reactions happen?
7. Why are they important to us?
8. How are displacement reactions explained as redox reactions?
Thank you!
Answer:
Acids react with most metals to form hydrogen gas and salt. ... When an acid reacts with metal, salt and hydrogen gas are produced
Preliminary preformulation studies for a new candidate drug molecule does not include one of these
A. Identity
B. Formula and weight
C. pharmacological activity
D. Pilot scale up
Answer:
Pilot scale up
Explanation:
Preformulation studies are carried out on candidate drug molecules that show sufficient pharmacological promise in animal model(pharma approach).
It involves preliminary study of the properties of a drug which is considered a potentially active ingredient against a particular disease condition.
Scale-up is the term used to refer to the increase in the batch size of a product. This is only done after a drug has been proven successful against the target disease after extensive pilot studies.
Scale-up is the last operation carried out when a drug has passed through all stages. It is not included in preliminary preformulation studies
What does the term "basic unit of matter" refer to?
O A.
Atoms
ОВ.
Elements
O c. Molecules
Explanation:
The term "basic unit of matter "refers to atom
A Atom
what is the IUPAC NAME FOR HPO4
Answer:
hydrogenphosphate I think
A quantity of 0.27 mole of neon is confined in a container at 2.50 atm and 298 Kand then allowed to expand adiabatically under two different conditions: (a) reversibly to 1.00 atm and (b) against a constant pressure of 1.00 atm. Calculate the final temperature in each case.
Answer:
a) Hence, T = 207 K.
b) Hence, T2 = 226 K.
Explanation:
Now the given,
n = 0.27 moles ; P = 2.5 atm ; T = 298 K
a) γ = 5/3 since Ne is a monoatomic gas.
[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]
Hence, T = 207 K
b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]
[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]
But P = P2
[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]
This gives us:
[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]
Hence, T2 = 226 K
Which process takes place when recharging a rechargeable battery?
a- Oxidation occurs at the positive anode.
b- Oxidation occurs at the positive cathode.
c- Oxidation occurs at the negative anode.
d- Oxidation occurs at the positive cathode.
Answer:
the answer is d.) potassium
Explanation:
Identify each of the following as a covalent compound or ionic compound. Then provide
either the formula for compounds identified by name or the name for those identified by
formula. (1 point each)
a. Li2O
b. Dinitrogen trioxide:
c. PCI3
d. Manganese(III) oxide:
Answer:
Explanation:
a) Ionic
Lithium oxide
b) Covalent
[tex]$\ce{N_2O_3}$[/tex]
c) Covalent
Phosphorus trichloride
d) Ionic
[tex]Mn_2O_3[/tex]
5) The properties of a substance depend on _______________
(a) the way ions are connected
(b) the ions it contains
(c) atoms
(d) the atoms it contains and the way these atoms are connected
Answer:
(d) the atoms it contains and the way these atoms are connected
Explanation:
hope it will be helpful for you
Explanation:
Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present. Physical properties can be measured without changing a substance's chemical identity.
Next, the chemist measures the volume of the unknown liquid as 0.610 L and the mass of the unknown liquid as 972. g.
Calculate the density of the liquid. Round
your answer to 3 significant digits.
1593.4 g / cm
10
Given the data above, is it possible to identify yes
Answer:
Density of liquid = 1.59 g/cm³
Explanation:
We'll begin by converting 0.610 L to cm³. This can be obtained as follow:
1 L = 1000 cm³
Therefore,
0.610 L = 0.610 L × 1000 cm³ / 1 L
0.610 L = 610 cm³
Finally, we shall determine the density of the liquid. This can be obtained as follow:
Volume of liquid = 610 cm³
Mass of liquid = 972 g
Density of liquid =?
Density = mass / volume
Density of liquid = 972 / 610
Density of liquid = 1.59 g/cm³
The density of any given liquid is equivalent to the mass of the liquid divided by the volume of the liquid.
From the given information, we have:
The mass of the unknown liquid to be = 972 g
The volume of the unknown liquid to be = 0.610 L = 610 cm³
If the formula for calculating [tex]\mathbf{Density = \dfrac{Mass}{Volume}}[/tex]
Then;
[tex]\mathbf{Density \ of \ the \ unknown \ liquid = \dfrac{972 \\ g}{610 \ cm}}[/tex]
[tex]\mathbf{Density \ of \ the \ unknown \ liquid = 1.593442623 \ g/cm}[/tex]
The Density of the unknown liquid ≅ 1.593 g/cm³
In conclusion, the density of the unknown liquid is 1.593 g/cm³ to 3 significant figures.
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A 4.17 L volume of oxygen gas measured at 7.62 °C is expanded to a new volume of 4.50 L. Calculate the temperature (in oC) of the gas at the higher volume, assuming no change in pressure.
Answer:
[tex]\boxed {\boxed {\sf 8.22 \ \textdegree C}}[/tex]
Explanation:
The question asks us to calculate the temperature of the gas at the higher volume. Since the pressure is constant, we are only concerned about volume and temperature. We will use Charles's Law. This states that the volume of a gas and the temperature of the gas have a directly proportionate relationship. The formula is:
[tex]\frac {V_1}{T_1}= \frac{V_2}{T_2}[/tex]
The gas starts at a volume of 4.17 liters and a temperature of 7.62 degrees Celsius.
[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{V_2}{T_2}[/tex]
The gas is expanded to a new volume of 4.50 liters, but the temperature is unknown.
[tex]\frac {4.17 \ L}{7.62 \textdegree C} = \frac{4.50 \ L}{T_2}[/tex]
We want to solve for the temperature at a higher volume. We must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.
[tex]4.17 \ L * T_2 = 4.50 \ L * 7.62 \textdegree C[/tex]
The variable is being multiplied by 4.17 liters. The inverse of multiplication is division. Divide both sides by 4.17 L.
[tex]\frac {4.17 \ L * T_2 }{4.17 \ L}= \frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]
[tex]T_2=\frac{4.50 \ L * 7.62 \textdegree C}{4.17 \ L}[/tex]
The units of liters (L) cancel.
[tex]T_2=\frac{4.50 * 7.62 \textdegree C}{4.17 }[/tex]
[tex]T_2=\frac{34.29}{4.17 } \textdegree C[/tex]
[tex]T_2=8.22302158273 \textdegree C[/tex]
The original measurements of liters and temperature have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.
The 3 to the right in the thousandth place tells us to leave the 2 in the hundredth place.
[tex]T_2 \approx 8.22 \ \textdegree C[/tex]
The temperature of the gas at the higher volume is approximately 8.22 degrees Celsius.
Different control mechanisms are used to regulate the synthesis of glycogen.
a. True
b. False
If 8.89 g of 2-methylcyclohexanone (112.17 g/mol) was reduced to 5.14 g of 2-methylcyclohexanol (114.19 g/mol), what is the percentage yield of the product?
Answer:
56.8%
Explanation:
The reaction of the problem is 1:1. That means 1 mole of 2-methylcyclohexanone produce 1 mole of 2-methylcyclohexanol.
Percentage yield is defined as 100 times the ratio between actual yield of the reaction (5.14g) and the theoretical yield.
The theoretical yield (All reactant produce products) is obtained from the mass of the reactant as follows:
Theoretical Yield:
8.89g 2-methylcyclohexanone * (1mol/112.17g) = 0.07925 moles 2-methylcyclohexanone
Assuming all reactant produce the product in a 100% of yield, the moles of 2-methylcyclohexanol are 0.07925 moles and the mass (Theoretical yield) is:
0.07925 moles 2-methylcyclohexanol * (114.19g/mol) = 9.05g
Percentage yield:
5.14g / 9.05g * 100 = 56.8%
The percentage of the mass successfully converted into a new product is 57.2%.
Mass of the reactantThe mass of the reactant (2-methylcyclohexanone) before the reduction is given as 8.89 g.
Mass of the product yieldedThe mass of the product ( 2-methylcyclohexanol) produced is given as 5.14 g.
Percentage yield of the productThe percentage of the mass successfully converted into a new product is calculated as follows;
[tex]= \frac{5.14}{8.99} \times 100\% \\\\= 57.2 \ \%[/tex]
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Write the formulas of all species in solution for the following ionic compounds by writing their dissolving equations:
(Use the lowest possible coefficients.)
1. Rubidium hydroxide: __--__+___
2. Sodium carbonate: __--__+__
3. Ammonium selenite:__--__+__
Answer:
1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
Explanation:
Let's consider the dissolving equations for the following compounds.
1. Rubidium hydroxide
RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Sodium carbonate
Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. Ammonium selenite
(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
