Answer:
Light is produced when an electron moves from a higher to a lower energy level. The color of the light (as we see it) depends on the amount of energy emitted by the electron.
Explanation:
According to the Bohr's model of the atom, energy is given out by an atom when an electron moves from a higher to a lower energy level. The energy of the emitted light is equal to the energy difference between the two energy levels.
Hence ;
∆E= E2 - E1
Where;
E2 = energy of the higher energy level
E1 = energy of the lower energy level.
By supplying light of various frequencies and observing the frequency/wavelength of light emitted by a particular atom, its electron configuration can be determined.
If under a given set of conditions the reaction A → B occurs with ΔG = -14 kJ/mol, and the reaction C→ B occurs with ΔG =- 16 kJ/mol, then:_______ a. Conversion of A to C is exergonic. b. A and C can never be at equilibrium, even under different reaction conditions. c. Oconversion of A to C is entropically driven. d. Conversion of C to A is freely reversible.
Answer:
a. Conversion of A to C is exergonic.
Explanation:
In the problem: C→ B occurs with ΔG = + 16 kJ/mol
It is possible to sum ΔG of reactions to obtain ΔG of another related reaction:
A → B ΔG = -14kJ/mol
C → B ΔG = +16kJ/mol
A → C ΔG = -14kJ/mol - (+16kJ/mol) = -30kJ/mol
As ΔG < 0
The reaction is exergonic
Under these condition, the reaction occurs. But under another conditions, the reaction will be at equilibrium (ΔG = 0)
A reaction is entropically driven if ΔG < 0 and ΔS is high. But we don't have information of ΔS.
As ΔG <0, the reaction is not spontaneous in the reverse direction
Right option is:
a. Conversion of A to C is exergonic.Do ions with more valence electrons become stable?
Answer:
yes!!!!!!!!!!
Explanation:
now go learn son
9. A Toyota Prius hybrid gets 21 kilometers per liter in highway driving. What is
the mileage in miles per gallon. (Given 1km = 0.621mi and 1L = 0.264gal)
Answer:
...................................................................................
The mileage of the Toyota Prius hybrid in miles per gallon would be 49.39 miles/gallon.
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.
As given in the problem a Toyota Prius hybrid gets 21 kilometers per liter in highway driving.
1 kilometers = 0.621 miles
21 kilometers = 21 × 0.621 miles
=13.04 miles
1 liter = 0.264 gallon
The mileage in miles per gallon = 13.04 miles / 0.264 gallon
= 49.39 miles/gallon
Thus, the mileage of the Toyota Prius hybrid in miles per gallon would be 49.39 miles/gallon.
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The speed of light in a vacuum is 2.998×108 m/s 2.998 × 10 8 m / s . Calculate its speed in miles per hour (miles/h m i l e s / h ).
Answer:
The speed of light in a vacuum is 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]
Explanation:
Two magnitudes are directly proportional when increasing one quantity increases the other in the same proportion or when decreasing one quantity decreases the other in the same proportion.
The rule of three allows the resolution of problems that are related to the proportionality of three known values and a fourth value that is always an unknown. In other words, it is useful to establish the proportionality between 2 values a and b through the knowledge of a third value c in order to calculate a fourth value x. In the case of direct margins, the rule of three between a, b and c and the unknown x is:
a ⇒ b
c ⇒ x
So: [tex]x=\frac{c*b}{a}[/tex]
In this case, knowing that 1 meter is equal to 0.000621 miles, 1 second is equal to 0.000278 hours, the simple rule of three is applied as follows: if 1 meter is equal to 0.000621 miles, 2.998 * 10⁸ meters are equal to how many miles?
[tex]miles=\frac{2.998*10^{8} meters*0.000621 miles}{1 meter}[/tex]
miles=186,175.8
[tex]2.998*10^{8} \frac{m}{s} =\frac{2.998*10^{8}m}{s}=\frac{186,175.8miles}{s}[/tex]
Replacing the seconds by their equivalent in meters:
[tex]\frac{186,175.8miles}{s} =\frac{186,175.8miles}{0.000278 hours}= 669,697,122.3 \frac{miles}{h}[/tex]
Then:
2.998*10⁸ [tex]\frac{m}{s}[/tex] = 669,697,122.3 [tex]\frac{miles}{h}[/tex] ≅ 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]
The speed of light in a vacuum is 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]
Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures are, respectively, 1.00 atm and 0.500 atm. When the system comes to equilibrium at 1100 K, the total pressure in the flask is found to be 1.35 atm. Given: 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔrH = − 198.2 kJ. mol-1 1.1 Calculate Kp at 1100 K
Answer:
The answer is "[tex]\bold{0.525\ \ atm^{-1}}[/tex]"
Explanation:
Given equation:
[tex]2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)[/tex]
Given value:
[tex]\Delta rH =-198.2 \ \ \frac{KJ}{mol}[/tex]
[tex]Kp=1100 \ K[/tex]
[tex]\Delta x = 2-(2+1)\\\\[/tex]
[tex]= 2-(2+1)\\\\= 2-(3)\\\\= -1[/tex]
[tex]\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right[/tex]
calculating the total pressure on equilibrium= [tex](1-2x)+(0.5-x)+2x \ atm\\\\[/tex]
[tex]= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\[/tex]
[tex]\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\[/tex]
calculating the pressure in [tex]So_2[/tex]:
[tex]= (1-2 \times 0.15)[/tex]
[tex]= 1-0.30 \\\\ =0.70 \ atm[/tex]
calculating the pressure in [tex]O_2[/tex]:
[tex]= (0.5- 0.15)\\\\= 0.35 \ atm \\[/tex]
calculating the pressure in [tex]So_3[/tex]:
[tex]= (2 \times 0.15)\\\\= (.30) \ atm \\\\[/tex]
Calculating the Kp at 1100 K:
[tex]= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\[/tex]
A 30ml sample of a liquid has a mass of 50 grams. What is the density of the liquid?
Answer:
Density = 1.67 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]Density = \frac{mass}{volume} [/tex]
From the question
mass = 50 g
volume = 30 mL
Substitute the values into the above formula and solve for the density
That's
[tex]Density = \frac{50}{30} \\ = \frac{5}{3} \\ = 1.66666...[/tex]
Wr have the final answer as
Density = 1.67 g/mLHope this helps you
A patient needs 40.0 mg of an antibiotic per kilogram of body weight each day. If the patient weighs 55 kilograms, how much antibiotic, in milligrams, should the patient receive each day for optimal therapy?
Answer:
2,200 milligrams
Explanation:
40 mg per 1 kilogram means 40 times 55 which is 2,200 mg
A 10.0 cm3 sample of copper has a mass of 89.6 g. What is the density of copper
Answer:
Density = 8.96 g/cm³Explanation:
The density of a substance can be found by using the formula
[tex]Density = \frac{mass}{volume} [/tex]
From the question
mass of copper = 89.6 g
volume = 10 cm³
Substitute the values into the above formula and solve
That's
[tex]Density = \frac{89.6}{10} [/tex]
We have the final answer as
Density = 8.96 g/cm³Hope this helps you
Look at the image of the solar system,
According to Kepler, which planet travels the fastest?
Neptune
Our Solar System
O Earth
Saturn
O Jupiter
O Mercury
Neptune
Mars
Uranus
Asteroid Bilt.
The Sun
Mercury
Earth
Venus
Jupiter
Mark this and return
Save and Exit
Nex
Submit
Answer:
Mercury
Explanation:
In one to two sentences, explain a similarity and a difference between the particles in liquid water at 100ºC and the particles in steam at 100ºC. PLEASSSSSSSSSSSSSSSE HELPPPPPPPPPPPPPPP ASAPPPPPPPPPP........................
WOW! I have the same question! Again lol.
A similarity is that both are at the same temperature and a difference is both are at different states of matter between the particles in liquid water at 100ºC and the particles in steam at 100ºC.
What are states of matter?There are only 3 states in which matter is present in the universe that is solid, liquid and gas, and on the basis of molecular space, they are differentiated in their respective category.
The water present both at 100ºC first is in a liquid state where molecules of water are more close as compared to the molecules of the steam water but the temperature for both is the same.
Therefore, the similarity is that both are at the same temperature and a difference is both are at different states of matter between the particles in liquid water at 100ºC and the particles in steam at 100ºC.
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What is the symbol for the ion that contains 12 protons, 10 electrons, and 12 neutrons?
A 2.600×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.9 mL . The density of water at 20.0∘C is 0.9982 g/mL.Part ACalculate the concentration of the glycerol solution in percent by mass.Express your answer to four significant figures and include the appropriate units.Part BCalculate the concentration of the glycerol solution in parts per million.Express your answer as an integer to four significant figures and include the appropriate units.
Answer:
A. 0.2395 w/w %
B. 2394ppm
Explanation:
A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:
Mass glycerol / Total mass * 100
Mass glycerol:
The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):
2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol
Mass of water:
998.9mL and density = 0.9982g/mL:
998.9mL * (0.9982g/mL) = 997.1g of water.
That means percent by mass is:
% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %
B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:
2.394g * (1000mg / 1g) = 2394mg:
Parts per million: 2394mg / L = 2394ppm
Considering the definition of percent by mass
A) the concentration of the glycerol solution in percent by mass is 0.2395%.
B) the concentration of the glycerol solution is 2394.34 ppm.
Concentration of the glycerol solution in percent by mass
A) The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.
The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
[tex]mass percentage=\frac{mass of solute}{mass of solution} x100[/tex]
In this case, you have a 2.600×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.
So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:
number of moles of glycerol= 2.600×10⁻² M× 1 L
number of moles of glycerol= 2.600×10⁻² moles
Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:
2.600×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.39434 grams of glycerol
On the other side, the volume of water needed was 998.9 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:
998.9 mL×0.9982 [tex]\frac{g}{mL}[/tex]= 997.1 grams of water
Finally, the mass percentage of the solution can be calculated as:
[tex]mass percentage=\frac{mass of glycerol}{mass of glycerol + mass of water} x100[/tex]
Solving:
[tex]mass percentage=\frac{2.39434 grams}{2.39434 grams+ 997.1 grams} x100[/tex]
[tex]mass percentage=\frac{2.39434 grams}{999.49434 grams} x100[/tex]
mass percentaje= 0.2395 %
In summary, the concentration of the glycerol solution in percent by mass is 0.2395%.
Parts per million (ppm)B) Parts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.
Being the mass of glycerol 2.39434 grams equal to 2394.34 mg (1 g=1000mg), the concentration is:
[tex]concentration=\frac{2394.34 mg}{1L}[/tex]
concentration= 2394.34 ppm
In summary, the concentration of the glycerol solution is 2394.34 ppm.
Learn more about:
mass percentage: brainly.com/question/19168984?referrer=searchResults brainly.com/question/18646836?referrer=searchResultsbrainly.com/question/24201923?referrer=searchResults brainly.com/question/9779410?referrer=searchResults brainly.com/question/17030163?referrer=searchResults ppmhttps://brainly.com/question/16727593?referrer=searchResultshttps://brainly.com/question/13565240?referrer=searchResultsA piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal? questions below A) 0.171 B) 0.343 C) 1.717 D) 3.433
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Explanation:
If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)
Answer:
[tex]m_{HNO_3}=27.4gHNO_3[/tex]
Explanation:
Hello,
In this case, for the chemical reaction:
[tex]3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(aq) + NO(g)[/tex]
The first step is to compute the theoretical yield of nitric acid via stoichiometry in terms of the 3:2 ratio between nitrogen dioxide (molar mass = 46 g/mol) and nitric acid (molar mass = 63 g/mol) respectively:
[tex]m_{HNO_3}^{theoretical}=40.0gNO_2*\frac{1molNO_2}{46gNO_2}* \frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3} \\\\m_{HNO_3}^{theoretical}=36.52gHNO_3[/tex]
Now, the actual amount is computed by taking into account the 75.0-% percent yield:
[tex]m_{HNO_3}=0.75*36.5gHNO_3\\\\m_{HNO_3}=27.4gHNO_3[/tex]
Best regards.
During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction (ii) occurs. (i) N2O4(l) 2 N2H4(l) 3 N2(g) 4 H2O(g) (ii) 2 N2O4(l) N2H4(l) 6 NO(g) 2 H2O(g) In one experiment 12.7 g of NO formed when 101.1 g of each reactant was used. What is the highest percent yield of N2 that can be expected
Answer:
Maximum expected yield = 87.2%
Explanation:
Equations of reactions:
Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)
Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)
Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol
In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.
101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄
N₂O₄ is the limiting reactant
101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂
In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄
12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce
mass of N₂O₄ used = 12.98 g
mass of N₂H₄ used = 2.26 g
mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g
mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g
In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂
88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄
N₂O₄ is the limiting reactant.
88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂
Percentage yield = (theoretical yield/actual yield) * 100%
Percentage yield = (40.23/46.15) * 100% = 87.2%
Therefore, maximum expected yield = 87.2%
The density of gold is 19.3 grams per milliliter. What would that be in pounds per cup?
Answer:
[tex]\rho =10.07\frac{lb}{cup}[/tex]
Explanation:
Hello,
In this case, given that 1 cup equals 236.588 mL and 1 pound equals 453.6 g, the required density in pounds per cup turns out:
[tex]\rho =19.3\frac{19.3g}{mL}*\frac{1lb}{453.6g}*\frac{236.588mL}{1cup}\\ \\\rho =10.07\frac{lb}{cup}[/tex]
Best regards.
Storing sugar as long chains for later use is an example of a(n) ____________ chemical reaction.
Answer:
Endothermic
Explanation:
Storing sugar for later use is an example of an endothermic reaction because that energy is being absorbed.
What would you call a linear alkane that contains 8 carbons?
Answer:
octane
Explanation:
heptane: an alkane with 7 C's
nonane: an alkane with 9 C's
hexane: an alkane with 6C's
The name of the linear alkane that contains 8 carbons is octane. The correct option is C.
What are alkanes?Saturated hydrocarbons are alkanes. This indicates that single bonds are used to connect each of their carbon atoms. They are not reactive. They react with oxygen, the process called combustion or burning. Examples are methane, ethane, etc.
Alkane is of many carbon chains. They can be single, double, triple, etc. The alkane that has 8 carbon atoms is called octane. A linear hydrocarbon is an octane. The bonds between each carbon in the molecular skeleton are limited to two.
5 carbon atoms are called pentane, six carbon atoms are called hexane, and seven carbon atoms are called heptane, as in series, the eighth carbon atom is called octane.
Thus, the correct option is C. octane.
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How many ml of 0.213-M Na3PO4 are required to deliver 66.4 mmol Na1+ ions?
Given :
Molarity of [tex]Na_3PO_4[/tex] , M = 0.213 M .
To Find :
How many ml of 0.213 M of [tex]Na_3PO_4[/tex] are required to deliver 66.4 mmol [tex]Na^+[/tex].
Solution :
Volume required :
[tex]V=[\text{ 66.4 mmol of }Na^+] + [\dfrac{\text{1 mm of }Na_3PO_4}{\text{3 mmol of }Na^+}]+[\dfrac{\text{1 ml }Na_3PO_4}{\text{0.213 mmol of }Na_3PO_4}][/tex]
So ,
[tex]V=\dfrac{66.4}{3\times 0.213}\ ml\\\\V=103.91\ ml[/tex]
Therefore , volume of [tex]Na_3PO_4[/tex] required is 103.91 ml .
Hence , this is the required solution .
What are likely formulas for the following molecules?
A) NH?OH.
B) AlCl?.
C) CF2Cl?.
D) CH?O.
Answer:
a
Explanation:
a the answers you welcome
The likely formulas for the molecules are NH₄OH, AlCl₃, CF₂Cl₂, CH₃O.
What is molecule ?According to the context, the term may or may not include ions that meet this requirement. A molecule is a collection of two or more atoms held together by the attractive forces known as chemical bonds.
A molecule is the smallest unit of any material that is composed of one or more elements and is capable of existing independently while maintaining all the substance's physical and chemical properties. Further atom division occurs within molecules. For instance, the symbols for the oxygen atom and molecule are O and O2, respectively.
The smallest identifiable unit into which a pure substance may be divided while retaining its composition and chemical properties is a molecule, which is a collection of two or more atoms.
Thus, The likely formulas for the molecules are NH₄OH, AlCl₃, CF₂Cl₂, CH₃O.
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A cylinder container can hold 2.45 L of water. It’s radius is 4.00 cm. What is the volume of it in cubic centimeters?
Answer:
2450 cm3
Explanation:
Volume of cylinder = V=πr2h
2.45L = 2450mL
1mL = 1 cm cubed
2450mL = 2450 cm cubed
The volume of the cylinder container in cubic centimeters is 2.45 × 10³ cm³.
We usually can find the volume of a cylinder applying the following expression.
[tex]V = \pi r^{2} h[/tex]
where,
r: radiush: heightSince we don't know the height, we will have to use the information that it can hold 2.45 L of water. We can convert liters to cubic centimeters using the following conversion factors:
1 L = 1000 mL1 mL = 1 cm³[tex]2.45 L \times \frac{1000mL}{1L} \times \frac{1cm^{3} }{1mL} =2.45 \times 10^{3} cm^{3}[/tex]
The volume of the cylinder container in cubic centimeters is 2.45 × 10³ cm³.
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What type(s) of intermolecular forces are expected between PF2Cl3 molecules?a. dispersion.b. dipole-dipole.c. ion-ion.d. hydrogen bonding.
Answer:
dispersion.
Explanation:
The molecule, PF2Cl3 is trigonal bipyramidal. The dipoles in the molecule cancel out since there is a symmetric charge distribution around the molecule hence the resultant dipole moment of the molecule is zero.
If the molecule is nonpolar, then the dominant intermolecular forces present are the weak dispersion forces, hence the answer above.
A 2 mole sample of F2(g) reacts with excess NaOH(aq) according to the equation above. If the reaction is repeated with excess NaOH(aq) but with 1 mole of F2(g), which of the following is correct?
Group of answer choices
The amount of OF2(g) produced is doubled.
The amount of OF2(g) produced is halved.
The amount of NaF(aq) produced remains the same.
The amount of NaF(aq) produced is doubled.
Answer:
B - The amount of OF2(g) produced is halved.
Explanation:
Because its right.
When instead of 2 moles of F₂, 1 mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.
Let's consider the balanced reaction that occurs when F₂ reacts with NaOH.
2 F₂(g) + 2 NaOH(aq) → OF₂(g) + 2 NaF(aq) + H₂O(l)
First, let's see the moles of OF₂ and NaF obtained from 2 moles of F₂, using the molar ratios derived from the balanced chemical equation.
[tex]2 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 1 mol OF_2\\\\2 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 2 mol NaF[/tex]
Now, let's compare with the moles of OF₂ and NaF obtained from 1 mol of F₂, again using the same molar ratios derived from the balanced chemical equation.
[tex]1 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 0.5 mol OF_2\\\\1 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 1 mol NaF[/tex]
As we can see, since we have half the amount of F₂, we obtain half the amount of the products. Then, the only right option is: The amount of OF₂(g) produced is halved.
When instead of 2 moles of F₂, 1 mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.
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Find the volume (in mL) of a substance that has a mass of 11.6 g and a density of 0.81 g/mL. Give your answer with two decimals.
Answer:
14.83mL
Explanation:
11.6/0.81=14.3
The weak acid HA is 2% ionized (dissociated) in a 0.20 M solution.
1. What is Ka for this acid?
2. What is the pH of this solution?
Answer:
1. Ka = 8.16x10⁻⁵
2. pH = 2.40
Explanation:
1. The dissociation of a weak acid in water occurs as follows:
HA ⇄ H⁺ + A⁻
Ka = [H⁺] [A⁻] / [HA]
As 2% of the 0.20M solution is dissociated:
[H⁺] = [A⁻] = 0.20M * 2% = 0.004M -As H⁺ and A⁻ comes from the same reaction, their concentrations are the same
[HA] = 0.20M * 98% = 0.196M
Ka = (0.004)² / (0.196M) = 8.16x10⁻⁵
2. pH = -log [H⁺] = -log [0.004M]
pH = 2.40SO2 can act as both oxidising and reducing agent
Answer:
No, it acts only as a reducing agent
Explanation:
Sulphur dioxide is a strong reducing agent. It is mainly used to prevent dried fruit from being oxidised and discoloured.
Which of the following substances has the lowest density?
A) A mass of 1.5 kg and a volume of 1.2 L
B) A mass of 25 g and a volume of 20 mL
C) A mass of 750 g and a volume of 70 dL
D) A mass of 5 mg and a volume of 25 UL (mcL)
Answer:
C) A mass of 750 g and a volume of 70 dL .
Explanation:
Hello,
In this case, for substantiating the substance having the lowest density we need to compute it in the same units for each case as shown below:
[tex]\rho=\frac{m}{V}[/tex]
A) [tex]\rho =\frac{1.5kg}{1.2L}*\frac{1000g}{1kg} *\frac{1L}{1000mL}=1.25g/mL[/tex]
B) [tex]\rho =\frac{25g}{20mL}*=1.25g/mL[/tex]
C) [tex]\rho =\frac{750g}{70dL}*\frac{10dL}{1L}*\frac{1L}{1000mL} =0.107g/mL[/tex]
D) [tex]\rho =\frac{15mg}{25\mu L}*\frac{1g}{1000mg} *\frac{1000\mu L}{1mL}=0.6g/mL[/tex]
Therefore, the lowest density corresponds to C) A mass of 750 g and a volume of 70 dL
Regards.
First two drop down menus choices:
1. treasury
2. ionic
3. matrimonial
4. covalent
Third drop down menu choices:
1. 40kJ/mol
2. 4000kJ/mol
3. 400kJ/mol
4. 40000kJ/mol
A medium-sized banana (118g) on average contains 422 mg of potassium - a nutrient needed to maintain fluid balance. If someone eats a medium-sized banana each day for a week, how many total grams of potassium are ingested in that week?
Answer:
2.954grams of pottasium for that week
Explanation:
According to this question, a medium-sized that weighs 118grams contains 422 milligrams (mg) of potassium nutrient.
If someone eats a medium-sized banana each day for a week, this means that the person has consumed 422mg of pottasium that day.
Since there are 7days in a week, the person will consume 422 × 7 = 2954 mg of pottasium that week.
The question asks for the total pottasium intake in grams for that week, hence, we need to convert 2954 milligrams to grams.
1000mg = 1 g
Hence, 2954mg will be 2954/1000
= 2.954grams
Therefore, 2.954grams of pottasium are ingested by that person that week.
student desing an experiment to see if the thermo energy is transfer at the same rate in different liquids the students safely heat 2 different room temperatures liquids on a hot plate an measure the temperature after 10 minutes what variable such the student control
Answer:
www.doe.virginia.gov/testing/sol/standards_docs/...
Explanation:
did you get the answer??