Answer:
Mole = molecular weight / molecular mass
Mole = 16/16
Mole= 1
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.
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
Different control mechanisms are used to regulate the synthesis of glycogen.
a. True
b. False
If the solvent front moves 8.0 cm and the two components in a sample being analyzed move 3.2 cm and 6.1 cm from the baseline, calculate the Rf values.
Answer:
Rf₁ = 0.40Rf₂ = 0.76Explanation:
We can calculate the Rf values by using the following formula:
Rf = Distance from the baseline / Solvent front distance
With that in mind we now proceed to calculate the Rf value for both components:
Rf₁ = 3.2 cm / 8.0 cm = 0.40Rf₂ = 6.1 cm / 8.0 cm = 0.76A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is ________ M-1s-1. A) 12
Answer: 0.085 (Ms)⁻¹
Explanation: Half life = 12 s
is the initial concentration = 0.98 M
Half life expression for second order kinetic is:
k = 0.085 (Ms)⁻¹
The rate constant for this reaction is 0.085 (Ms)⁻¹ .
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
sự sắp xếp nguyên tử trong vật chất
Answer:
sosksjsjjs
Explanation:
even i know how to type şïllily
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]
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
In 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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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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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..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:
A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment, she recovers 2.775 g of sand and 0.852 g of salt.a. What was the percent composition of sand in the mixture according to the student's data? b. What was the percent recovery?
Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
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.
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.
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
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
The following compound can be identified as
Answer:
3: Lactone
Explanation:
Lactones are defined as carboxylic esters that contain the structure (−C(=O)−O−) which is essentially showing that an ester has now become part of the chemical structure of the ring.
Thus, looking at the question, it has the structure as defined in Lactones.
Thus, we can say that the compound is a Lactone.
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.
When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net ionic equation for the reaction.
Answer:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:
[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]
Now, we can separate the nitrates in ions as they are aqueous to obtain:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]
And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Best regards!
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
Identify the compound in the following group that is most soluble in water. Match the words in the left column to the appropriate blanks in the sentences on the right. ResetHelp Of the three compounds butanoic acid, butane, and butanone, the one that is most soluble in water is . This is because its functional group can form the intermolecular forces with polar water.
butanoic acid strongest hydrocarbon butanone butane alcohol carboxylic acid What carboxylic acid is found in each of the following substances? Drag the appropriate descriptions to their respective bins. Reset rancid butter stinging red ants Methanoic acid Butanoic acid Propanoic acid Ethanoic acid Review Co Draw the structure of methyl butanoate, Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms toolbars. H: 129 uxo com H o с + N 1 0 S a CH Br р
Answer:
Following are the responses to the given points:
Explanation:
For question 1:
Butanoic acid, butane, and butanone are also the three chemicals most dissolve in water. Its intermolecular force forces are produced by carboxylic acid functional groups with water.
For question 2:
Butanoic acid is a rancid buffer.
Methanoic acid is responsible for the stinging red ants
For question 3:
Methyl butanoate's chemical structure.
What was the plum pudding atomic model
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)
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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For the following reaction, 15.4 grams of chlorine gas are allowed to react with 49.6 grams of sodium iodide. chlorine (g) sodium iodide (s) sodium chloride (s) iodine (s) What is the maximum amount of sodium chloride that can be formed
Answer:
19.3 g of NaCl can be produced
Explanation:
We state the reaction:
Cl₂ (g) + 2NaI (s) → 2NaCl (s) + I₂ (s)
We need to determine limiting reagent:
15.4 g . 1mol /70.9g = 0.217 moles of chlorine
49.6 g . 1mol / 149.89g = 0.331 moles of NaI
Ratio is 1:2. 1 mol of chlorine reacts to 2 moles of NaI
0.217 moles may react to (0.217 . 2)/1 = 0.434 moles of NaI
It is ok to say the NaI is the limting reactant because we need 0.434 moles of it and we only have 0.331.
Ratio is 2:2.
0.331 moles of NaI can produce 0.331 moles of NaCl
We convert mass to moles: 0.331 mol . 58.45g /mol = 19.3 g
what is the hybridisation of the central carbon in CH3C triple bonded to N
Explanation:
the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet
which of these statements is true about planets? Planets
A. revolve around the sun
B. are spherical in shape
C. rotate in its axis
D. all of the above
Answer:
D. all of the above
Explanation:
A and C are verified by Keplar's laws of planetary motion.
B is verified by the equatorial and polar aces of the Planet.
Answer:
c
Explanation:
they dont have to orbit the sun specifically and are commonly more ovoid than spherical
Consider the following data on some weak acids and weak bases:
Acid Base Ca
Name Formula Name Formula
Hydrocyanic acid HCN 4.9 x 10^-10 Ammonia NH3 1.8x 10^-5
Hypochlorous acid HCIO 3.0x10^-8 Ethylamine C2H5NH2 6.4 x 10^-4
Use this data to rank the following solutions in order of increasing pH.
Solution pH
0.1 M NaCN
0.1M C2H5NH3Br
0.1 M Nal
0.1 M KCIO
Answer:
0.1 M Nal
0.1M C2H5NH3Br
0.1M KClO
0.1M NaCN
Explanation:
The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:
In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).
The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:
0.1 M Nal
0.1M C2H5NH3Br
0.1M KClO
0.1M NaCN
