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]
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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repining of fruits is which type of change
Answer:
irreversible.
I hope this will help you
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
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.
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)
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
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.
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!
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.
The 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
Which of the following is not organic compound?
a. CH4
b. H2CO3
c. CCl4
d. CH3-OH
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.
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 = 0.76If 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!
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
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 % .
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
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.
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 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]
Different control mechanisms are used to regulate the synthesis of glycogen.
a. True
b. False
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.
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..sự sắp xếp nguyên tử trong vật chất
Answer:
sosksjsjjs
Explanation:
even i know how to type şïllily
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
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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 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
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.
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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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
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