Answer:
b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.
Explanation:
The Bohr effect is a phenomenon described by Christian Bohr. Is an affinity that binds oxygen and hemoglobin and is inversely related to the concentration of carbon dioxide. As CO2 reacts with water and an increase in CO2 results in a decrease in blood ph.For a given fluorophore, select the choice that correctly lists the processes of fluorescence, absorption, and phosphorescence in order from shortest to longest wavelength.
a. absorption, fluorescence, phosphorescence
b. Fluorescence = phosphorescence, absorption
c. fluorescence, phosphorescence, absorption
d. phosphorescence, fluorescence, absorption
e. absorption, phosphorescence, fluorescence
f. absorption, fluorescence = phosphorescence
Answer:
absorption, fluorescence = phosphorescence
Explanation:
Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.
Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.
The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.
The combustion of hydrogen-oxygen mixtures is used to produce very high temperatures (ca. 2500 °C) needed for certain types of welding operations. Consider the reaction to be
H₂(g) + ½O₂(g) → H₂O(g) ∆H° = -241.8 kJ/mol
What is the quantity of heat evolved, in kilojoules, when a 180 g mixture containing equal parts of H₂ and O₂ by mass is burned?
in kj
Answer:
1360kJ are evolved
Explanation:
When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.
To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:
Moles H2 -Molar mass: 2g/mol-
90g H2 * (1mol / 2g) = 45 moles
Moles O2 -Molar mass: 32g/mol-
90g * (1mol / 32g) = 2.81moles
For a complete reaction of 2.81 moles of O2 are needed:
2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2
As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.
As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:
2.81 moles O2 * (241.8kJ / 1/2moles O2) =
1360kJ are evolvedThe quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is
The equation for the combustion of hydrogen is given as:
[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]
Recall that:
number of moles = mass/molar mass:Since the mass of 180 g is equally shared by H₂ and O₂, then:
mass of H₂ = 90 gmass of O₂ = 90 gThe number of moles of the reactant can be determined as follows:
For H₂:
[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]
no of moles = 44.6 mol
For O₂:
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]
no of moles = 2.8 mol
Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:
If 1/2 moles of O₂ produces -241.8 kJ/mol of water;
Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:
[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]
[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]
= - 1358.91 kJ
≅ - 1360 kJ
Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ
Learn more about the quantity of heat here:
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examples s name of thosse food items we can store for a month?
Answer:
1. Nuts
2. Canned meats and seafood
3. Dried grains
4. Dark chocolate
5. Protein powders
Given the chemical equation: KI +Pb(NO3)2—>KNO3 + Pbl2
Balance this chemical equation.
Indicate the type of reaction. How do you know?
Thoroughly discuss how your balanced chemical equation agrees with the law of conservation of mass.
Answer:
[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]
Double replacement reaction.
It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).
Explanation:
Hello there!
In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:
[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]
Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).
Regards!
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 192 grams of phosphorus pentabromide to molecules.
Convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules.
Answer:
1) 2.69 * 10²³ PBr₅
2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁
Explanation:
Question 1)
We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.
Phosphorus pentabromide is given by PBr₅.
To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.
To convert from grams to moles, we will find the molar mass of PBr₅.
Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:
[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]
And since we want to convert from grams to moles, we can write the following ratio:
[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]
Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.
To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.
So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
In combination, starting with 192 grams of PBr₅, we will acquire:
[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
Cancel like units:
[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]
Multiply. Hence:
[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]
Since the final answer should have three significant digits, our final answer is:
[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]
So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.
Question 2)
We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.
3.42 kilograms is equivalent to 3420 grams of table sugar.
Again, we can convert from grams to moles, and then from moles to molecules.
First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:
[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]
To cancel units, we can write our ratio as:
[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]
With grams in the denominator.
And by definition:
[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Combining the two ratios and the starting value, we acquire:
[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Cancel like units:
[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
Rewrite:
[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
The resulting answer should have three significant digits. Hence:
[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]
So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.
Answer:
2.69×10²³ molecules of PBr₅
6.02×10²⁴ molecules of C₁₂H₂₂O₁₁
Explanation:
To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.
[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]
Now, we know that there are about 2.69×10²³ molecules of PBr₅.
To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.
[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]
Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.
Write the equation for the reaction between sulfuric acid solution and solid aluminum hydroxide. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)
Explanation:
Let's consider the balanced chemical equation that takes place when sulfuric acid solution and solid aluminum hydroxide react to form aluminum sulfate and water. This is a neutralization equation.
3 H₂SO₄(aq) + 2 Al(OH)₃(s) ⇒ Al₂(SO₄)₃(aq) + 6 H₂O(l)
what is Lewis acid and Lewis base? give examples
Explanation:
example is copper iron...........
Part A
If the theoretical yield of a reaction is 23.5 g and the actual yield is 14.8 g, what is the percent yield?
Answer:
[tex]\boxed {\boxed {\sf 63.0 \%}}[/tex]
Explanation:
The percent yield is the ratio of the actual yield to the theoretical yield.
[tex]percent \ yield = \frac{actual \ yield}{theoretical \ yield} * 100[/tex]
The actual yield is the amount obtained from performing a chemical reaction. For this problem, it is 14.8 grams. The theoretical yield is the potential amount from performing a chemical reaction at maximum performance. For this problem, it is 23.5 grams.We can substitute the known values into the formula.
[tex]percent \ yield= \frac{ 14.8 \ g}{23.5 \ g}*100[/tex]
Divide.
[tex]percent \ yield = 0.629787234043*100[/tex]
Multiply.
[tex]percent \ yield = 62.9787234043[/tex]
The original measurements for the theoretical and actual yields have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place.
The 7 to the right, in the hundredths place, tells us to round the 9 up to a 0. Since we rounded up to 0, we have to move to the next place to the left and round the 2 up to a 3.
[tex]percent \ yield \approx 63.0[/tex]
The percent yield is approximately 63.0 percent.
Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcohol with two other carbons attached to the carbon with the hydroxyl group_____.An alcohol with one other carbon attached to the carbon with the hydroxyl group____.An alcohol with three other carbons attached to the carbon with the hydroxyl group____.
Answer:
1). 1-pentanol - Primary
2). 3-ethyl-3-pentanol - Tertiary
3). 2-hexanol - Secondary
4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - Secondary
5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - Primary
6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - Tertiary
Explanation:
The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.
In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.
In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.
In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example: 3-ethyl-3-pentanol, -tert -butyl alcohol, etc.
Curium – 245 is an alpha emitter. Write the equation for the nuclear reaction and identify the product nucleus.
Answer:
Please find the complete solution in the attached file.
Explanation:
Which of the following is not organic compound?
a. CH4
b. H2CO3
c. CCl4
d. CH3-OH
The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chloride dissolves in water. Be sure to specify states such as (aq) or (s).
Answer:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Explanation:
Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
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.
Question 2 10
10 Points
Which of the following chemical equations depicts a balanced chemical equation?
O A. AgNO, Kcro > KNO, Agro,
OB. AgNO3 + Kycro, » 2K NO; + Agro,
C.3AgNO3 + 2K,Cro--> 3KNO3 + 249900,
D. 2AgNO, K Cro-> 2KNO; 1Cro,
Resol Selection
Answer:
2AgNO, K Cro-> 2KNO; 1Cro,
Choose the correct answer to make the statement true.
a. An exothermic reaction has a positive ΔH and absorbs heat from the surroundings.
b. An exothermic reaction feels warm to the touch. a positive ΔH and gives off heat to the surroundings.
c. An exothermic reaction feels warm to the touch. a negative ΔH and absorbs heat from the surroundings.
d. An exothermic reaction feels warm to the touch. a negative ΔH and gives off heat to the surroundings.
e. An exothermic reaction feels warm to the touch.
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!
3) Convert 0.250 moles of aluminum sulfate to grams.
4) Convert 2.70 grams of ammonia to moles.
Answer:
0.000731 grams aluminium sulfate
46.0 mols ammonia
Explanation:
ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol
[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]
NH3 has a molar mass of 17.031 g/mol
[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]
Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]
we have to find the 0.250 moles of aluminum sulphate.
[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]
[tex]\\\\\\[/tex]
Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]
We have to find 2.70 grams of ammonia
[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]
An example of a molecular compound that obeys the octet rule in which all atoms have a zero formal charge is:
Answer:
[tex]NCl_3[/tex]
Explanation:
An octet rule is a thumb rule in the chemical sciences in which there is a natural tendency for an atom to prefer eight electrons in the valence shell of the atom. When there are less than eight electrons in the atom, they react with other atoms and form more stable compounds.
In the context, nitrogen trichloride, [tex]NCl_3[/tex], is an example of molecular compound which obeys the octet rule having a zero formal charges on each atom of the compound.
Hi,Valency of chlorine is 1. Why?Thank you
Answer:
The chlorine element belongs to group 17 because it has 7 valence electron . Its valency is 1 . It can gain one electron from any other atom to become stable. This means that it can never form a double or triple bond.
How many molecules of Iron(II)oxide are present in 35.2*10^-23 g of Iron (II)oxide?
Answer:
R.F.M of Iron (II) oxide :
[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]
Moles :
[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]
Molecules :
[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]
The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.
What is Avogadro's number?Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.
The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.
Given, the mass of the iron oxide = 35.2 ×10⁻²³ g
The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol
71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³
35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)
The number of molecules of FeO in a given mass = 2.95 molecules
Learn more about Avogadro's number, here:
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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
A voltaic cell is constructed with an Ag/Ag half-cell and a Pb/Pb2 half-cell. The silver electrode is positive. Write the balanced half-reactions and the overall reaction. Include the phases of each reactant and product.
Answer:
Following are the chemical equation to the given question:
Explanation:
The Electrode is a silver film that is covered with such a thin coating of silver chloride, either by dipping its wire directly into silver-molten chloride, plating the wire using hydrogen peroxide, or oxidation silver in a chloride. In the given silver electrode, this anode acts as a cathode and thus reduces.
Half of the response reduction: [tex]Ag^+(aq)+e^-\rightarrow Ag(s)[/tex]
Half-effect oxidation: [tex]Pb(s)\rightarrow Pb^{2+}(aq)+2e^-[/tex]
Complete reaction: [tex]Pb(s)+2Ag^+(aq) \rightarrow Pb^{2+}(aq)+2Ag(s)[/tex]
repining of fruits is which type of change
Answer:
irreversible.
I hope this will help you
If the specific heat of methanol is 32.91 J/K. g. how many joules are necessary to raise the temperature of 120 g of methanol from 24 0C to 98 0C?
Answer:
[tex]Q=292240.8J=292.2kJ[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to use the general heat equation:
[tex]Q=mC(T_2-T_1)[/tex]
For us to plug the given mass, specific heat and temperature change to obtain the required heat:
[tex]Q=120g*32.91\frac{J}{g*K} (98\°C-24\°C)\\\\Q=292240.8J=292.2kJ[/tex]
Regards!
3. Which of the following statements is not correct?
A. All hybridization must involve an s-orbital
B. Excitation of Carbon atom in CH, involves promotion of one is electron to the empty 2p-orbital
C. Hybridization in transition elements can take the form dsp*
D. Hybridization involves only the valence electrons and orbitals
E. None of the above
Answer:
Excitation of Carbon atom in CH, involves promotion of one is electron to the empty 2p-orbital
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
300g de acido comercial se disuelve en agua destilada contenidos en un cono, cuyo radio es de 0.005Km y 300cm de altura, si la densidades de 1.2g/m3 ¿Cuál es la concentración expresada en %m/m?
Answer:
97.0%m/m es la concentración de la solución
Explanation:
El porcentaje masa/masa (%m/m) se define como 100 veces el radio entre la masa de soluto (300g de HCl) y la masa de la solución. Para hallar la masa de la solución debemos hallar la masa del agua (Solvente) haciendo uso de la ecuación del volumen de un cono. Con el volumen del cono podemos hallar la masa del agua haciendo uso de la densidad, así:
Volumen:
Volumen Cono = π*r²*h / 3
Donde r es el radio = 0.300m
h la altura = 5m
Volumen = π*(5m)²*0.300m / 3
7.85m³
Masa Agua:
7.85m³ * (1.2g / m³) = 9.42g Agua
Masa solución:
300g HCl + 9.42g Agua = 309.42g Solución
%m/m:
300g HCl / 309.42g * 100 =
97.0%m/m es la concentración de la soluciónIf 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 = 0.76The mass of a crucible and lid is 23.422 g. After adding a sample of hydrate compound the crucible, cover, and contents weigh 24.746 g. After heating with a Bunsen burner to remove the water of hydration, the mass of the crucible, cover, and sample was 24.213 g. How many moles of water did the hydrate compound contain
Answer:
0.030 mole
Explanation:
Mass of crucible + lid = 23.422 g
Mass of crucible + lid + compound = 24.746 g
Mass of crucible + lid + compound - water = 24.213
Mass of water = Mass of crucible + lid + compound + heat
= 24.746 - 24.213
= 0.533 g
Mole of water in the hydrated compound = mass of water in the compound/molar mass of water
= 0.533/18
= 0.0296 mole = 0.030 mole
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 % .
If you reacted 88.9 g of ammonia with excess oxygen, what mass of water would you expect to make? You will need to balance the equation first.
NH3(g) + O2(g) -> NO(g) + H2O(g)
Explanation:
here's the answer to your question
