Answer:
The amount of solute that will dissolve
Explanation:
A P E X
A buffered solution has a pH of 7.5. What would happen to the pH if a small
amount of acid were added?
Answer:
Dear user,
Answer to your query is provided below
When small amount of acid was added to buffered solution, pH will change very less.
Explanation:
Buffer solution resists change in ph on adding small amount of acid or base but when we calculate the value of buffer capacity we take the change in ph when we add acid or base to 1 lit solution of buffer.This contradicts the definition of buffer solution.
Write the chemical reaction for hydrogen thiocyanate in water, whose equilibrium constant is Ka. Include the physical states for each species
Write the chemical reaction for thiocyanate ion in water, whose equilibrium constant is Kb. Include the physical states for each species
Answer:
HSCN (aq) + H₂O(l) ⇄ SCN⁻(aq) + H₃O⁺(aq) Ka
SCN⁻ (aq) + H₂O(l) ⇄ HSCN (aq) + OH⁻(aq) Kb
Explanation:
We identify the formula:
HSCN → hydrogen thiocyanate which is also known as Thiocyanic acid
HSCN (aq) + H₂O(l) ⇄ SCN⁻(aq) + H₃O⁺(aq) Ka
As an acid, it gives proton to the solution. It is a weak acid, because the Ka
Ka = [SCN⁻] . [H₃O⁺] / [HSCN]
As a weak acid, the thiocyanate ion, will be the conjugate strong base. In water It can make hydrolisis:
SCN⁻ (aq) + H₂O(l) ⇄ HSCN (aq) + OH⁻(aq) Kb
As a base, it takes a proton from water.
Kb = [HSCN] . [OH⁻] / [SCN⁻]
H-S-C-N is the structure of hydro-thi-ocyanate.
H-S-C-N structure:1. combines with water, it produces the ion thiocyanate and the ion hydronium.
2. When thiocyanate combines with water, it produces the ion hydrogen thiocyanate and the ion hydroxy.
Find out more information about 'Thiocyanate'.
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A sample of carbon dioxide gas (CO2) contains 6 x 1022 molecules. How many moles of carbon dioxide does this represent?
Answer:
Explanation:
1 ST METHOD
GIVEN DATA:
number of molecules=6.022×10²²molecules
Avogodro's number Na=6.022×10²²
TO FIND:
number of moles
SOLUTION:
As we knom that number of moles=number of molecules/Na
number of moles=6.022×10²²/6.022×10²²
number of moles=1 moles
2ND METHOD:
AS WE KNOW THAT A MOLE OF A SUBSTANCE CONTAINS 6.022×10²² PARTICLES OF THAT SUBATANCE SO
1MOLE OF CO2=6.022×10²²MOLECULES OF CO2
For the reaction A+B↽−−⇀C+D, assume that the standard change in free energy has a positive value. Changing the conditions of the reaction can alter the value of the change in free energy (ΔG). Classify the conditions as to whether each would decrease the value of ΔG, increase the value of ΔG, or not change the value of ΔG for the reaction. For each change, assume that the other variables are kept constant.
A. Adding a catalystb.
B. Increasing [C] and [D]
C. Coupling with ATP hydrolysis
D. Increasing [A] and [B]
Explanation:
a. Adding a catalyst
no effect .( Catalyst can only change the activation energy but not the free energy).
b. increasing [C] and [D]
Increase the free energy .
c. Coupling with ATP hydrolysis
decrease the free energy value .
d.Increasing [A] and [B]
decrease the free energy.
Suppose you are titrating a sulfuric acid solution of unknown concentration with a sodium hydroxide solution according to the equation H 2 S O 4 + 2 N a O H ⟶ 2 H 2 O + N a 2 S O 4 If you require 28.07 mL of 0.697 M NaOH solution to titrate 220.1 mL of H 2 SO 4 solution, what is the concentration of the H 2 SO 4 solution? Type answer:
Answer:
[tex]M_{acid}=0.0444M[/tex]
Explanation:
Hello,
In this case, for the given reaction:
[tex]H _2 S O _4 + 2 N a O H \rightarrow 2 _H 2 O + N a _2 S O_ 4[/tex]
We find a 1:2 molar ratio between the acid and the base respectively, for that reason, at the equivalence point we find:
[tex]2*n_{acid}=n_{base}[/tex]
That in terms of concentrations and volumes we can compute the concentration of the acid solution:
[tex]2*M_{acid}V_{acid}=M_{base}V_{base}\\\\M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}}=\frac{0.697M*28.07mL}{2*220.1mL}\\ \\M_{acid}=0.0444M[/tex]
Best regards.
The concentration of the sulfuric acid solution (H₂SO₄) is 0.0444 M
From the question,
We are to determine the concentration of the H₂SO₄ solution
Using the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
Where
[tex]C_{A}[/tex] is the concentration of acid
[tex]C_{B}[/tex] is the concentration of base
[tex]V_{A}[/tex] is the volume of acid
[tex]V_{B}[/tex] is the volume of base
[tex]n_{A}[/tex] is the mole ratio of acid
and [tex]n_{B}[/tex] is the mole ratio of base
The given balanced chemical equation for the reaction is
H₂SO₄ + 2NaOH ⟶ 2H₂O + Na₂SO₄
∴ [tex]n_{A} = 1[/tex]
and [tex]n_{B} = 2[/tex]
From the given information
[tex]V_{A} = 220.1 \ mL[/tex]
[tex]C_{B} = 0.697 \ M[/tex]
[tex]V_{B} = 28.07 \ mL[/tex]
Putting the parameters into the equation, we get
[tex]\frac{C_{A} \times 220.1}{0.697 \times 28.07}= \frac{1}{2}[/tex]
∴ [tex]C_{A} = \frac{1 \times 0.697 \times28.07}{2\times 220.1}[/tex]
[tex]C_{A} = \frac{19.56479}{440.2}[/tex]
[tex]C_{A} = 0.0444 \ M[/tex]
Hence, the concentration of the sulfuric acid solution (H₂SO₄) is 0.0444 M
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In all measurements, you are looking for values that are accurate. Pretend that a pebble with a known mass of 0.567 g was massed four times where the following masses were obtained: 0.256 g, 0.723 g, 0.554 g and 0.354 g. Those masses are an example of:__________.
a) being precise but not accurate.
b) being both precise and accurate.
c) being neither precise nor accurate.
d) being accurate but not precise.
Answer:
The correct answer is option c, that is, being neither precise nor accurate.
Explanation:
Obtaining the experimental values that come almost close to the true value is termed as accuracy. On the other hand, precision is obtaining experimental values continuously, which may come either far away or near to the true value.
Based on the given information, it is clear that 0.554 grams, that is, the third value obtained from the experiment is showing some closeness to the true value, which is 0.567 grams. While, neither any of the other values are close or is coming near to the true value. This shows that the measurements are an illustration of being neither precise nor accurate.
A sample of gas has a volume of 571 mL at a pressure of 4.04 atm. The gas is compressed and now has a pressure of 7.17 atm. Predict whether the new volume is greater or less than the initial volume, and calculate the new volume. Assume temperature is constant and no gas escaped from the container.
Answer:
The new volume is less than the initial volume.
The new volume is 322mL
Explanation:
Based on Boyle's law, the pressure of a gas is inversely proportional to its volume under constant temperature. That means if the pressure of a gas is increased, the volume decrease and vice versa. The formula is:
P₁V₁ = P₂V₂
Where P is pressure and V is volume of 1, initial state and 2, final states.
In the problem, the pressure of the gas increased from 4.04atm to 7.17atm, That means the new volume is less than initial volume because the gas is compressed occupying less volume.
Replacing in Boyle's equation:
4.04atm*571mL = 7.17atmV₂
322mL = V₂Beeing the new volume of the compressed gas 322mL
The pressure of sulfur dioxide (SO2) is 2.27 x 104 Pa. There are 418 moles of this gas in a volume of 57.9 m3. Find the translational rms speed of the sulfur dioxide molecules.
Answer:
The translational rms speed of the sulfur dioxide molecules. vrms = 52.8 m/s
Explanation:
The root-mean-square speed is the measure of the speed of particles in a gas. It is given by the formula below;
vrms=√3RT/M
where vrms is the root-mean-square of the velocity, M is the molar mass of the gas in kilograms per mole, R is the molar gas constant, and T is the temperature in Kelvin.
Molar mass of SO₂ in Kg/ mole = (64/1000) Kg/mol = 0.064 kg/mol
R = 8.314 m³.Pa/K*mol
T = ?
Using PV = nRT to find T
T = PV/nR
Substituting for T in the rms formula
vrms = √(3R*PV/nR)/M
vrms = √3PV/nM
vrms = √3 * 2.27 * 10⁴ * 57.9)/418 * 0.064
vrms = 52.8 m/s
DATA AND CALCULATIONS: (you must show your calculations) Part I. Determination of accuracy of a graduated cylinder Calculations: Experimental Step Measurable Mass of empty graduated cylinder 47.229 g Mass of filled graduated cylinder 71.821 g Mass of water (filled – empty) g Volume of water, calculated (calculated from mass of water, using the equation “density = mass/volume”, given the fact that the density of water is exactly 1 g/mL) mL Volume of water, measured (from the reading of the scale on the graduated cylinder) 25.0 mL Percent difference between measured and calculated volumes of water [(measured-calculated)/calculated] ×100% %
Answer:
[tex]\large \boxed{2 \, \%}[/tex]
Explanation:
1. Data
Mass of graduated cylinder = 47.229 g
Mass of graduated cylinder + water = 71.821 g
Actual volume of water = 25.0 mL
2. Calculations
(a) Mass of water
Mass = 71.821 g -47.229 g = 24.592 g
(b) Volume of water
[tex]\text{Volume} = \dfrac{\text{mass}}{\text{volume }} = \dfrac{\text{24.592 g}}{\text{ 1 g/mL}} = \text{24.592 mL}[/tex]
(c) Percent Difference
[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{Measured - Calculated}\lvert}{ \text{Calculated}} \times 100 \,\%\\\\& = & \dfrac{\lvert 25.0 - 24.492\lvert}{24.492} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.5\lvert}{24.492} \times 100 \, \%\\ \\& = & 0.02 \times 100 \, \%\\& = & \mathbf{2 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{2 \, \%} }$}[/tex]
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Explain why someone would choose to use octane in a car engine rather than octene?
A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.582 g CO2 and 3.922 g H2O. What percent by mass of oxygen is contained in the original sample?
Answer:
[tex]\% O=27.6\%[/tex]
Explanation:
Hello,
In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:
- Moles of carbon are contained in the 9.582 grams of carbon dioxide:
[tex]n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2} =0.218molC[/tex]
- Moles of hydrogen are contained in the 3.922 grams of water:
[tex]n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH[/tex]
- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:
[tex]m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO[/tex]
Finally, we compute the percent by mass of oxygen:
[tex]\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%[/tex]
Regards.
Upon the addition of water, As2O3 is converted to H3AsO3. During the titration H3AsO3 is oxidized to H3AsO4 and MnO4- is reduced to Mn2 . Write a balanced net ionic equation for the reaction
Answer:
5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)
Explanation:
Every net balanced ionic equation is composed of a union of two half equations;
The oxidation half equation (indicating electron loss) and the reduction half equation (indicating electron gain). Remember that redox reactions is a process in which electrons are lost and gained by chemical species simultaneously. One specie looses electrons in the oxidation half equation while the other specie gains electrons in the reduction half equation.
The balanced redox reaction equation shows the overall redox process and shows at a glance the total number of elect tribe lost or gained in the redox process. The overall redox reaction equation for the titration described in the question is;
5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)
A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units.
Answer:
25.0 grams is the mass of the steel bar.Explanation:
Heat gained by steel bar will be equal to heat lost by the water
[tex]Q_1=-Q_2[/tex]
Mass of steel= [tex]m_1[/tex]
Specific heat capacity of steel = [tex]c_1=0.452 J/g^oC[/tex]
Initial temperature of the steel = [tex]T_1=2.00^oC[/tex]
Final temperature of the steel = [tex]T_2=T=21.50^oC[/tex]
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Mass of water= [tex]m_2= 105 g[/tex]
Specific heat capacity of water=[tex]c_2=4.18 J/g^oC[/tex]
Initial temperature of the water = [tex]T_3=22.00^oC[/tex]
Final temperature of water = [tex]T_2=T=21.50^oC[/tex]
[tex]Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))[/tex]
On substituting all values:
[tex](m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}[/tex]
25.0 grams is the mass of the steel bar.Answer:
[tex]m_{steel}=24.9g[/tex]
Explanation:
Hello,
In this case, since the water is initially hot, the released heat by it is gained by the steel rod since it is initially cold which in energetic terms is illustrated by:
[tex]\Delta H_{water}=-\Delta H_{steel}[/tex]
That in terms of mass, specific heat and temperature change is:
[tex]m_{water}Cp_{water}(T_f-T_{water})=-m_{steel}Cp_{steel}(T_f-T_{steel})[/tex]
Thus, we simply solve for the mass of the steel rod:
[tex]m_{steel}=\frac{m_{water}Cp_{water}(T_f-T_{water})}{-Cp_{steel}(T_f-T_{steel})} \\\\m_{steel}=\frac{105mL*\frac{1g}{1mL}*4.18\frac{J}{g\°C}*(21.50-22.00)\°C}{-0.452\frac{J}{g\°C}*(21.50-2.00)\°C} \\\\m_{steel}=24.9g[/tex]
Best regards.
Carry out the following operations as if they were calculations of experimental results and express each answer in standard notation with the correct number of significant figures and wtih the correct units. Provide both the answer and the units.
1. 5.6792 m + .6 m + 4.33 m
2. 3.70 g - 2.9133 g
3. 4.51 cm x 3.6666 cm
Answer:
1. [tex]10.6\; \rm m[/tex] (one decimal place.)
2.[tex]0.79\; \rm g[/tex] (two decimal places.)
3. [tex]16.5\;\rm cm^2[/tex] (three significant figures.)
Explanation:
1.The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.
For example, in the first expression:
[tex]5.6792\;\rm m[/tex] has four decimal places.[tex]0.6\; \rm m[/tex] has only one decimal place.[tex]4.33\; \rm m[/tex] has two decimal places.Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: [tex]\rm m[/tex].)
Therefore:
[tex]\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m[/tex]. (Rounded to one decimal place.)
2.Similarly:
[tex]\rm 3.70\; \rm g[/tex] has two decimal places.[tex]2.9133\; \rm g[/tex] has four decimal places.Therefore, the result should be rounded to two decimal places. Its unit should be [tex]\rm g[/tex] (same as the unit of the two inputs.)
[tex]\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g[/tex]. (Rounded to two decimal places.)
3.When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:
[tex]4.51\; \rm cm[/tex] has three significant figures.[tex]3.6666\; \rm cm[/tex] has five significant figures.Therefore, the result should have only three significant figures.
The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be [tex]\rm cm \cdot cm[/tex], which is occasionally written as [tex]\rm cm^2[/tex].
[tex]\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2[/tex]. (Rounded to three significant figures.)
Question 1 of 10
How is each element represented in the periodic table?
A. Each element is in an order based on alphabetical order.
B. Each element is listed as an abbreviation of the first letters of its
DE ME
C. Each element is represented by a one or two-letter symbol.
D. Each element is listed in its own box based on when it was
discovered
Answer:
C
Explanation:
A- incorrect, not in alphabetical order
B- incorrect, symbol for salt in NA not SA
C- correct
D- incorrect, not based on discovery
Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.
What is Periodic Table ?
The periodic table, also known as the periodic table of the elements, is a tabular display of the chemical elements.
It is widely used in chemistry, physics, and other sciences, and is generally seen as an icon of chemistry.
Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.
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Temperature on Reaction Rate Use the drop-down menus to answer the questions. Which form of the sodium bicarbonate tablet has the most surface area? As the surface area increases, what happens to the average time required for the reaction?
Answer:Crushed, decreased
Explanation:
Just got it right
A solution has a hydrogen ion (or hydronium ion) concentration of 1.00×10−9 M.
What is the pH of the solution?
7.5
8.0
8.5
9.0
what is the maximum number of electrons in p&q shell
Answer:
6 electrons
Explanation:
:)
hope this helps :))))))))
Answer:
each p shell can hold max. 6 electorns
i dont know what a q shell is
How does the A Hreaction relate to the A He of molecules involved in a reaction?
A. A Hreaction = A Hf, reactants
B. AHreaction = A Hf, products - AHf, reactants
C. A Hreaction = AHf, products
D. A Hreaction = A Ht, products + AHf, reactants
Answer:
B. ΔHreaction = ΔH°f reactants- ΔH°f products
Explanation:
Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.
Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).
For example, in ΔH°f of H₂O, the equation is:
H₂(g) + 1/2O₂(g) → H₂O(g)
The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).
Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:
NaOH + HCl → H₂O + NaCl. ΔHr
The ΔH°f for each substance in the reaction is:
NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)
HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)
H₂O: H₂ + 1/2O₂ → H₂O (3)
NaCl: Na + 1/2Cl₂ → NaCl (4)
The algebraic sum of (3) + (4) is -(ΔH°f reactants):
H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants
This reaction - {(1)+(2)} ΔH°f products
NaOH + HCl → H₂O + NaCl.
ΔHr = ΔH°f reactants- ΔH°f products
In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:
B. ΔHreaction = ΔH°f reactants- ΔH°f productsQuestion
2 Points
An object in a fluid will sink if:
A. the buoyant force is larger than the weight of the object.
B. the buoyant force is larger than the mass of the object.
C. the buoyant force is smaller than the weight of the object.
D. the buoyant force is smaller than the mass of the object.
SUBM
Answer:
c
Explanation:
when the weight of the object is greater than the buoyant force sinking is occurred.
Which of the following is a correct representation of the isotope of sulfur that has 19 neutrons? sulfur-19 sulfur-35 sulfur-16 sulfur-32
Answer:
sulfur-35
Explanation:
Sulfur-35 is a radioactive isotope that contains 19 neutrons.
Isotopes are represented with mass numbers. Mass number is the addition of number of proton and number of neutrons.
The number of proton in sulfur = 16
Number of neutron = 19
So, mass number = no. of protons + no. of neutrons
= 16 + 19
= 35
Hence, the correct answer is sulfur-35.
Answer:
sulfur -35
sulfur negative thirty fiveArrange the ff substances in order of increasing tendency to donate electrons with reasons
Alpha- ketoglutarate, Oxaloacetate ,O2, NADP+, Malate
Answer:
Alpha- ketoglutarate --> NADP+ --> Oxaloacetate --> O2.
(O2 has the HIGHEST tendency to donate electrons).
Explanation:
Okay, we are to Arrange Alpha- ketoglutarate, Oxaloacetate ,O2, NADP+ and Malate in order of increasing tendency to donate electrons, therefore, the correct order is;
Alpha- ketoglutarate --> NADP+ --> Oxaloacetate --> O2.
And this result can be gotten if we consider these compounds in their oxidative mechanism and check their potential values.
(1). When Alpha- ketoglutarate react with CO2, it gives isocitrate and standard redox potential value of -0.38.
(2). NADP+ with react with two moles of electrons and one mole of hydrogen ion to produce NADPH and standard redox potential value of -0.32.
(3). Oxaloacetate will react with two moles of hydrogen ion and electrons respectively to produce standard redox potential value of -0.18 and Malate.
(4). O2 will react with 4 moles of hydrogen ion and electrons respectively to produce 2 moles of water and standard redox potential value of +0.82.
CONCLUSION: the higher the negative values of the potential, the lesser the electron affinity and vice versa.
Which image best represents the particles in liquids
Answer:
The 2nd Picture represents the particles in liquids.
Explanation:
A volume of 38.7 mL of H2O is initially at 28.0 oC. A chilled glass marble weighing 4.00 g with a heat capacity of 3.52 J/oC is placed in the water. If the final temperature of the system is 26.1 oC , what was the initial temperature of the marble? Water has a density of 1.00 g/mL and a specific heat of 4.18 J/goC. Enter your answer numerically, to three significant figures and in terms of oC.
Answer:
- 61.2°C = Initial T°
Explanation:
This is a calorimetry problem, where the heat from the water was gained by the marble.
Q = m . C . ΔT
where ΔT is final T° - initial T°, C is the specific heat and m, mass.
By the volume of water, we realize the mass (we apply density):
1 g/mL = mass / 38.7 mL
38.7 g = mass of water
Now, we need to find out the specific heat for the marble and we have Heat Capacity data
Heat Capacity = C . m
3.52 J/°C / 4 g = C → specific heat → 0.88 J/g °C
We make the equations for both heats:
m . C . ΔT from water = m . C . ΔT from the marble
38.7 g . 4.18 J/g°C ( 26.1°C - 28°C) = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)
307.35 J = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)
- 307.35 is a negative value, because the was has decreased the temperature, is a loss of heat, but we have to work with the positive number.
307.35 J / (4 . 0.88 °C/J) = 26.1°C - Initial T°
87.32°C = 26.1°C - Initial T°
- 61.2°C = Initial T°
Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds
The given question is incomplete, the complete question is:
Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?
Answer:
The lowest whole-number mass ratio in the two compounds is 1:2.
Explanation:
There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.
The no. of moles can be determined by using the formula,
moles = mass/molecular mass
moles = 2.98 g/207.2 g/mol
= 0.0144 moles
The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.
moles = 0.461 g /16 g/mol
= 0.0288 moles
The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2
On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,
moles = 9.89 g / 207.2 g/mol
= 0.0477 moles
The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,
moles = 0.763 g / 16 g
= 0.0477 moles
The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1
Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.
Choose the species that is incorrectly matched with its electronic geometry.
1. BeBr2 : linear
2. CF4 : tetrahedral
3. NH3 : tetrahedral
4. H2O : tetrahedral
5. PF3 : trigonal bipyramidal
Answer:
PF3 : trigonal bipyramidal
Explanation:
PF3 has 4 domains around the central phosphorus (3 shared pairs and one lone pair of electrons), thus the electron geometry that has 4 domains is tetrahedral not trigonal bipyramidal
From the options the specie that is incorrectly matched is ( 5 ) ; PF₃ : trigonal bipyramidal
The specie PF₃ is composed of 3 shared pairs and one unshared pair of electrons ( i.e. It has 4 domains ) as seen in the Lewis structure of PF₃. therefore when writing its electronic geometry, it should expressed/written as tetrahedral and not trigonal bipyramidal.
Hence we can conclude that The specie that is incorrectly matched is PF3 : trigonal bipyramidal
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Arrange the compounds in order of increasing acid strength. Consult the references section for pKa values of each acid. Suppose the strongest acid is b, the acid of intermediate strength is c and the weakest acid is a Then this leads to the following input:
a. CH 3NH3
Methylammonium ion
b. CH3 COH
Acetic acid
c. H2S
Hydrogen sulfide
The strongest acid is b, the acid of intermediate strength is c and the weakest acid is a Then this leads to the H2S (Hydrogen sulfide). Therefore, option C is correct.
What do you mean by an acid ?A chemical that, when combined with certain metals, emits hydrogen ions in water and forms salts. Acids have a sour taste and cause certain dyes to turn red.
An acid is a substance that releases hydrogen ions into solution, whereas a base or alkali absorbs hydrogen ions.
In an aqueous solution or water, a weak acid is an acid that has been partially dissociated into its ions. A strong acid, on the other hand, completely dissociates into its ions in water. The pH of weak acids is higher than that of strong acids.
Thus, The strongest acid is b, the acid of intermediate strength is c and the weakest acid is a Then this leads to the H2S, option C is correct.
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Why are covalent substances gases and liquid rather than solids?
Covalent compounds are held together with an intra molecular attraction which is weaker than metallic bond
hence covalent compounds exist as liquids, gases and soft solids
Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g. How many grams of fat will you gain if you eat 10,000 kJ (about 2390 kcal) one day and do nothing but sit relaxed for 14.3 h and sleep for the other 9.70 h
Answer:
[tex]m=107.8g[/tex]
Explanation:
Hello,
In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h of sleeping:
[tex]E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ[/tex]
Then, we compute the energy used that day:
[tex]E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ[/tex]
Finally, the mass by considering the consumed fat:
[tex]m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g[/tex]
Regards.
Express your answer to three significant figures.
This balanced equation shows the reaction of sodium hydroxide and sulfuric acid:
2NaOH + H2SO4 - Na2SO4 + 2H20.
In a laboratory experiment, a student mixes 355 grams of sulfuric acid with an excess of sodium hydroxide. What is the theoretical mass of
sodium sulfate produced? Refer to the periodic table and the polyatomic ion resource.
The theoretical mass of sodium sulfate is
grams.
Answer: The theoretical mass of [tex]Na_2SO_4[/tex] is, 514 grams.
Explanation : Given,
Mass of [tex]H_2SO_4[/tex] = 355 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
First we have to calculate the moles of [tex]H_2SO_4[/tex].
[tex]\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}[/tex]
[tex]\text{Moles of }H_2SO_4=\frac{355g}{98g/mol}=3.62mol[/tex]
Now we have to calculate the moles of [tex]Na_2SO_4[/tex]
The balanced chemical equation is:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react to give 1 mole of [tex]Na_2SO_4[/tex]
So, 3.62 mole of [tex]H_2SO_4[/tex] react to give 3.62 mole of [tex]Na_2SO_4[/tex]
Now we have to calculate the mass of [tex]Na_2SO_4[/tex]
[tex]\text{ Mass of }Na_2SO_4=\text{ Moles of }Na_2SO_4\times \text{ Molar mass of }Na_2SO_4[/tex]
Molar mass of [tex]Na_2SO_4[/tex] = 142 g/mole
[tex]\text{ Mass of }Na_2SO_4=(3.62moles)\times (142g/mole)=514g[/tex]
Therefore, the theoretical mass of [tex]Na_2SO_4[/tex] is, 514 grams.