Answer:
Most positive = rubidium
Most negative = fluorine
Explanation:
Electron affinity of an element is the energy released when an electron is attached to a neural atom to form an ion in its gaseous state.
X + e⁻ → X⁻
Electron affinity is similar to electronegativity which is the tendency at which an atom accepts an ion towards itself.
Electron affinity increases across the period and decreases down the group in the periodic table.
In the above option,
Fluorine has the highest electron affinity
Rubidium has the lowest electron affinity
Tellurium and then finally Phosphorus
Helium in this case would have the lowest electron affinity because it has filled orbital and does not require any electron to attain stability. Technically, Helium has the lowest or is expected to have the lowest electron affinity which is close to zero according to quantum mechanics.
Most positive = rubidium
Most negative = fluorine.
You can check periodic table for their exact values
What is the change in energy, in kJ, when 45.3 grams of methanol, CH3OH, combusts? 2\text{CH}_3\text{OH}\left(l\right) + 3\text{O}_2\left(g\right)\rightarrow2\text{CO}_2\left(g\right)+4\text{H}_2\text{O}\left(g\right)\hskip .5in \Delta\text{H}=-726\text{ kJ}2 CH 3 OH ( l ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 4 H 2 O ( g ) Δ H = − 726 kJ Group of answer choices -513 kJ +2,050 kJ -1,030 kJ -2,050 kJ +513 kJ
Answer: -1,030 kJ
Explanation:
To calculate the number of moles we use the equation:
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar mass}}[/tex] .....(1)
Putting values in equation 1, we get:
[tex]\text{Moles of methanol}=\frac{45.3g}{32g/mol}=1.42mol[/tex]
The balanced chemical reaction is:
[tex]CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex] [tex]\Delta H=-726kJ[/tex]
Given :
Energy released when 1 mole of [tex]CH_3OH[/tex] combusts = 726k J
Thus Energy released when 1.42 moles of [tex]CH_3OH[/tex] combusts = [tex]\frac{726kJ}{1}\times 1.42=1030J[/tex]
Thus 1030 kJ of heat is released and as [tex]\Delta H[/tex] is negative for exothermic reaction, [tex]\Delta H=-1030kJ[/tex]
What is the mole ratio of water to H3PO4?
Answer:
Explanation:
Phosphoric acid H₃PO₄ is produced from reaction of water and tetraphosphorus decoxide P₄O₁₀ as follows
P₄O₁₀ + 6 H₂O = 4 H₃PO₄
In this reaction 6 molecules of H₂O and one molecule of phosphorus compound P₄O₁₀ is needed to produce phosphoric acid , ie the conversion factor of water to acid is 6 / 1
This ratio is called mole ratio of water to H₃PO₄.
So the required ratio is 6 : 1 .
Description (with words) of water just above melting temperature. What intermolecular forces do you expect to find in water in liquid state
Answer:
intermolecular dipole-dipole hydrogen bonds
Explanation:
Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.
Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.
A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is saying 91g I keep getting 175g. Can someone please explain?
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
Calculate the internal energy of 2 moles of argon gas (assuming ideal behavior) at 298 K. Suggest two ways to increase its internal energy by 10 J.
The internal energy of 2 moles of argon gas (Ar) at 298 K is 186.07 J.
By increasing the no.of moles and temperature the internal energy can be increased.
What is internal energy ?The internal energy of a system is its overall energy associated with its random motions. Thermodynamically it is a state function because the change in its values depends only on the initial and final states of the system and does not depend upon the path.
For an ideal gas, the internal energy is calculated using the following equation:
U = n C T
here, n is the no.of moles, C be the molar heat capacity at constant volume and T be the temperature in K. C for Ar is 0.3122 J/(Kg K) and temperature and no.of moles are given 298 K and 2 moles respectively.
Thus U = 2 mols × 0.3122 J/ (Kg K) × 298 K
= 186.07 J
By increasing the amount of gas and heating the system, the internal energy can be increased. By increasing the temperature by 20 K will raise the internal energy by 10 J.
Hence, by these two ways the internal energy of a gaseous system can be increased.
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how do you fight off ADHD medication
Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.
Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.
It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ksp for Ag₂CO₃ is 8.10 × 10⁻¹²)
Answer:
[tex]\large \boxed{1.64\times 10^{-5}\text{ mol/L }}[/tex]
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
[tex]K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}[/tex]
The amount of the sample in space is referred to as concentration, in this the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].
Concentration Calculation:In chemistry and related sciences, the phrase "concentration" is frequently used. It is a metric for determining how much of one material was mixed with the other.The solution's concentration is indeed the amount of solute absorbed in a given quantity of liquid or solution, following are the calculation of the concentration of [tex]Ag^+[/tex]:Concentration of [tex]Na_2CO_3 = 0.00750 M[/tex]
[tex](CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)\\\\[/tex]
So, the Maximum concentration of [tex]Ag^+[/tex] is [tex](3.28 \times 10^{-5}\ M)[/tex].
Find out the more information about the concentration here:
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Which best describes the act of using senses or tools to gather information? creating a hypothesis making an observation summarizing the results recording the measurements
Answer:
B - Making an Observation
Explanation:
Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.
What are senses in the scientific method?The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.
Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.
The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.
Thus, option B is correct.
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Will sodium iodide react with bromine to produce sodium bromide and iodine? Why or why not
A.Yes, because bromine has lower activity than iodine
B.Yes, because bromine has higher activity than iodine
C.No, because bromine has lower activity than iodine
D.No,because bromine has higher activity than iodine
Answer:
C.No, because bromine has lower activity than iodine
Explanation:
Hello,
In this case, for answering the given question we should remember that the higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, we can say that the described reaction is not possible due to the fact that bromine is less active. For that reason answer is C.No, because bromine has lower activity than iodine.
Best regards.
Answer:
Yes, because bromine has lower activity than iodine
Explanation:
Chromic acid, H2CrO4, is a strong oxidizing agent that oxidizes primary and secondary alcohols into carboxylic acids and ketones, respectively. Which technique would be most useful for monitoring the progress of the reaction
Answer:
Thin-layer chromatography
Explanation:
Here in this question, we are concerned with giving the best technique that could be use for monitoring the progress of the reaction that involves the oxidation of alcohols.
The most useful technique here for monitoring the progress of the reaction would be the thin layer chromatography.
Firstly, we should know the reason why we are using a chromatographic technique. This is intuitive. The reason is that the oxidizing agent being used is a colored substance. So basically, the progress of the oxidation reaction would solely rest on the fact that we have a decrease in color saturation of the said oxidizing agent.
This is best done using the thin layer chromatography because we have a mobile phase that moves quickly over the stationary phase and asides this, it moves evenly.
This makes the fact that color changes can be quickly and easily identified and we can tell if the oxidation reaction has gone to completion or not
a) If you weigh a hydrated compound before and after heating it. Do you expect the mass to be
more or less than before heating? Explain why.
b) How about if you weigh a bag of popcorn before and after heating it. Does the popped corn
weigh more or less than the unpopped kernels? Explain why.
Answer:
less becuase that moisture adds weight
Explanation:
For the following reaction, 142 grams of silver nitrate are allowed to react with 22.3 grams of copper . silver nitrate(aq) copper(s) copper(II) nitrate(aq) silver(s) What is the maximum amount of copper(II) nitrate that can be formed
Answer:
even I have the same dought
What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?
Answer:
8.33mL or .0083L
Explanation:
Use m1 * V1 = m2 * V2
6.00M(x) = 0.100M(500mL)
solve for x
x= (.1 * 500) / 6
x=8.333 mL
Turn on Write equation. What you see is an equation that shows the original uranium atom on the left. The boxes on the right represent the daughter product—the atom produced by radioactive decay—and the emitted alpha particle.
Answer:
Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.
Explanation:
Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:
[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]
I hope this explanation is clear and explanatory.
A chemical reaction has the equation 2AgNO3 (aq) + Zn (s) 2Ag (s) + Zn(NO3)2 (aq). What type of reaction occurs between AgNO3 and Zn?
Answer: single replacement reaction
Explanation:
A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.
A general single displacement reaction can be represented as :
[tex]X+YZ\rightarrow XZ+Y[/tex]
The reaction [tex]2AgNO_3(aq)+Zn(s)\rightarrow 2Ag(s)+Zn(NO_3)_2(aq)[/tex]
When zinc metal is added to aqueous silver nitrate, zinc being more reactive than silver displaces silver atom from its salt solution and lead to formation of zinc nitrate and silver metal.
Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N 2O 4 and 45.0 g N 2H 4. Some possibly useful molar masses are as follows: N 2O 4 = 92.02 g/mol, N 2H 4 = 32.05 g/mol.
N 2O 4( l) + 2 N 2H 4( l) → 3 N 2( g) + 4 H 2O( g)
a) LR = N2O4, 45.7 g N2 formed
b) LR = N2O4, 105 g N2 formed
c) LR = N2H4, 13.3 g N2 formed
d) LR = N2H4, 59.0 g N2 formed
e) No LR, 45.0 g N2 formed
Answer:
Option A. LR = N2O4, 45.7g N2 formed
Explanation:
The balanced equation for the reaction is given below:
N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)
Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
Molar mass of N2 = 2x14.01 = 28.02g/mol
Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g
Summary:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4.
Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.
From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.
Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.
Finally, we shall determine the mass of N2 produced from the reaction.
In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.
The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:
From the balanced equation above,
92.02g of N2O4 reacted to produce 84.06g of N2.
Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.
Therefore, 45.7g of N2 were produced from the reaction.
At the end of the day,
The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.
Hydrogen peroxide decomposes to form water and oxygen gas according to the following equation: 2H2O2(aq) 2H2O(l) + O2(g) If 315 g of hydrogen peroxide, H2O2, decomposes and all the O2 gas is collected in a balloon at 0.792 atm and 23C, what is the volume of the O2 gas collected?
Answer:
141.89 dm^3
Explanation:
The equation of the reaction is;
2H2O2(aq) --------->2H2O(l) + O2(g)
Now , we are told that the mass of hydrogen peroxide decomposed was 315g. Number of moles of hydrogen peroxide in 315g of the substance is given by;
Number of moles= mass/molar mass
Molar mass of hydrogen peroxide= 34.0147 g/mol
Number of moles= 315g/34.0147 g/mol = 9.26 moles of hydrogen hydrogen peroxide.
From the reaction equation;
2 moles of hydrogen peroxide yields 1 mole of oxygen
9.26 moles of hydrogen peroxide yields 9.26 ×1/2 = 4.63 moles of oxygen
From the ideal gas equation;
Volume of the gas V= the unknown
Pressure of the gas P= 0.792 atm
Temperature of the gas= 23°C +273 = 296 K
Number of moles of oxygen = 4.63 moles of oxygen
R= 0.082atmdm^3K-1mol-1
Hence, from PV=nRT
V= nRT/P
V= 4.63 × 0.082 × 296/0.792 = 141.89 dm^3
Calculate the number of grams of sodium in 2.00 g
of each sodium-containing food additive.
-NaCl
-Na3PO4
-NaC7H5O2
-Na2C6H6O7
Answer:
A = 0.7871g
B = 0.8417g
C = 0.3192g
D = 0.3897g
Explanation:
Hello,
The number of grams of sodium (Na) in 2g of NaCl is
Molar mass of NaCl = 58.44g/mol
Molar mass of Na = 23g
23g of Na = 58.44g of NaCl
x g of Na = 2g of NaCl
x = (2 × 23) / 58.44
x = 0.7871g
Therefore, 0.7871g of Na is present in 2g of NaCl
2.
2g of Na₃PO₄
Molar mass of Na₃PO₄ = 163.94g/mol
Molar mass of Na = 23g
(3 × 23)g of Na = 163.94 g of Na₃PO₄
x g of Na = 2g of Na₃PO₄
x = (2 × 69) / 163.94
x = 0.8417g
0.8417g of Na is present in 2g of Na₃PO₄
3.
2g of NaC₇H₅O₂
Molar mass of NaC₇H₅O₂ = 144.103g/mol
Molar mass of Na = 23g
23g of Na = 144.103g of NaC₇H₅O₂
x g of Na = 2g of NaC₇H₅O₂
x = (2 × 23) / 144.103
x = 0.3192g
0.3192g of Na is present in 2g of NaC₇H₅O₂
4. 2g of Na₂C₆H₆O₇
Molar mass of Na₂C₆H₆O₇ = 236.08g/mol
Molar mass of Na = 23g
(2 × 23)g of Na = 236.08g of Na₂C₆H₆O₇
x g = 2g of Na₂C₆H₆O₇
x = (46 × 2) / 236.08
x = 0.3897g
0.3897g is present in 2g of Na₂C₆H₆O₇
A pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20
Answer:
C.
Explanation:
Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.
Select correct answer. The percent yield (isolation yield) of guaifenesin isolated from a 650 mg tablet containing 400 mg dose of drug, can be expressed as: Group of answer choices
Answer:
61.5%.
Explanation:
We usually define percentage yield in chemistry as follows;
Percentage yield= actual yield/theoretical yield ×100
In this case we have;
Actual yield of the guaifenesin 400mg
Theoretical yield of the guaifenesin 650 mg
Hence;
Percentage yield of the guaifenesin =
400/650 ×100 = 61.5%
Hence the theoretical yield of the guaifenesin is 61.5%.
Can a catalyst change an exothermic reaction into an endothermic reaction or vice versa? Please explain your answer.
Answer:
A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.
Explanation:
Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.
Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other releases it.
As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.
what is the calculated value of the cell potential at 298 k for an eleectrochemical cell with the following reaction, when the Cl2 pressure is 1.31 atm, the cl- concentration is at
HERE IS THE COMPLETE QUESTION
what is the calculated value of the cell potential at 298 k for an electrochemical cell with the following reaction, when the Cl2 pressure is 1.31 atm, the cl- concentration is at 1.28M, and the Ni2+ concentration is 1.24M
[tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]
Answer:
[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]
Explanation:
From the given question;
We can see that Nickel is oxidized and Chlorine is reduced. We also known that Oxidation occurs at the anode while reduction occurs at the cathode.
SO the Oxidation and the reduction reaction can be expressed as shown below.
Oxidation : [tex]Ni_{s} \to Ni ^{2+}_{aq} + 2e^- \ \ \ \ E^0 = -0.26[/tex]
Reduction : [tex]Cl_{2(g)} + 2e^- \to 2Cl^- _{(aq)} \ \ \ \ E^0} = 1.36[/tex]
[tex]Cl_{2(g)} + Ni _{(s)} \to 2Cl^-_{aq} + Ni^{2+}_{aq}[/tex]
[tex]E^0_{cell} = E^0_{cathode}-E^0_{anode}[/tex]
[tex]E^0_{cell} = E^0_{Cl_2/Cl^-}-E^0_{Ni^{2+}/Ni}[/tex]
[tex]E^0_{cell} = +1.36 - (-0.26)[/tex]
[tex]E^0_{cell} = +1.62 \ V[/tex]
By applying Nernst Equation; we have :
[tex]E_{cell} = E^0_{cell}- (\dfrac{0.0591}{n}) \ log Q[/tex]
where
n= 2 moles
[tex]Q= \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]
[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[Cl^-]^2[Ni^{2+}]}{P_{Cl_2}}[/tex]
[tex]E_{cell} = +1.62 \ V}- (\dfrac{0.0591}{2}) \ log \dfrac{[1.28]^2[1.24]}{1.31*10^{-4}}[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384][1.24]}{1.31*10^{-4}}[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[1.6384*1.24]}{1.31}*10^{4}[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) \ log \dfrac{[2.031616]}{1.31}*10^{4}[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085*10^{4})[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ log (1.55085)+ log \ 10^{4}[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 0.1905 + 4[/tex]
[tex]E_{cell} = +1.62 \ V}- (0.02955) * \ 4.1905[/tex]
[tex]E_{cell} = +1.62 \ V}- 0.1238[/tex]
[tex]\mathbf{E_{cell} = +1.4962 \ V}[/tex]
Each unknown mixture contains 5 metal constituents. Select the 5 metal ions that you have identified as being present in your mixture. Please double check your selections before you hit the submit button. a. Ca b. Co c. Cr d. Fe e. K f. Mn g. Zn
Explanation:
A metal ion is a type of atom compound that has an electric charge.
Such atoms willingly lose electrons in order to build positive ions called cations. The selected Ions are :
[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]
If a flask maintained at 674 K contains 0.138 moles of NH4I(s) in equilibrium with 4.34×10-2 M NH3(g) and 9.39×10-2 M HI(g), what is the value of the equilbrium constant at 674 K?
Answer:
K = 4.07x10⁻³
Explanation:
Based on the reaction:
NH₄I(s) ⇄ NH₃(g) + HI(g)
You can define K of equilibrium as the ratio of concentrations of reactants and products, thus:
K = [NH₃] [HI] / [NH₄I]
But, as NH₄I is a solid, is not taken into account in the equilibrium, that means K expression is:
K = [NH₃] [HI]
As the concentrations in equilibrium of the gases is:
[NH₃] = 4.34x10⁻²M
[HI] = 9.39x10⁻²M
Equilibrium constant, K, is:
K = 4.34x10⁻²M * 9.39x10⁻²M
K = 4.07x10⁻³Which of the following changes would improve the accuracy of the molar mass measurement carried out in this experiment:
a. use a solvent with a lower freezing point
b. use a solvent with a larger Kf
c. use a more concentrated solution
Sodium metal reacts with chlorine gas in a combination reaction. Write a balanced equation to describe this reaction. Click in the answer box to open the symbol palette. Include states of matter in your answer.
Answer:
Na + Cl ⇒ NaCl
Explanation:
Skeleton equation: Na + Cl ⇒ NaCl
This is already balanced so that is the answer.
a sample of gas at fixed pressure has a temperature of 300 k and a volume of 3 L, calculate the volume if the gas is heated to 700 k
Answer:
7L
Explanation:
using charles law
V1/T1 = V1/T1
Answer:
7l
Explanation:
What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? K f for water is 1.86°C/m. What is the freezing point of a solution of 7.15 g MgCl2 in 100 g of water? K f for water is 1.86°C/m. -0.140°C -2.80°C -1.40°C -4.18°C
Answer:
THE NEW FREEZING POINT IS -4.196 °C
Explanation:
ΔTf = 1 Kf m
molarity of MgCl2:
Molar mass = (24 + 35.5 *2) g/mol
molar mass = 95 g/mol
7.15 g of MgCl2 in 100 g of water
7.15 g = 100 g
(7.15 * 100 / 1000) = 1000 g or 1 L or 1 dm3
= 0.715 g /dm3
Molarity in mol/dm3 = molarity in g/dm3 / molar mass
= 0.715 g /dm3 / 95 g/mol
m = 0.00752 mol/ dm3
So therefore:
ΔTf = i Kf m
1 = 3 (1 Mg and 2 Cl)
Kf = 1.86 °C/m
M = 0.752 moles
So we have:
ΔTf = 3 * 1.86 * 0.752
ΔTf = 4.196 °C
The new freezing point therefore will be 0 °C - 4.196 °C which is equals to - 4.196 °C
When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.
Required:
a. What is the empirical formula of the compound?
b. How many grams of O2 were used in the reaction?
c. How many grams would have been required for complete combustion?
Answer:
(a) The empirical formula of the compound is
m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).
(b) The grams of O2 that were used in the reaction is 1.146 g
(c) The amount of O2 that would have been required for complete combustion is 1.401 g.
Explanation:
a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)
(b) Using law of conservation of mass from above
m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)
m(O2) = 0.446 + 0.700 + 0.430 - 0.430
m(O2) = 1.146 g
The grams of O2 that were used in the reaction is 1.146 g
(c) for complete combustion, we need to oxidized CO to CO2
Then, 2CO +O2 = 2CO2
m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}
m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g
Note; Molar mass of O2 = 32, CO = 28
m(total)(O2) = m(O2) + m(add)(O2)
m(total)(O2) = 1.146 + 0.255 = 1.401 g
The amount of that grams would have been required for complete combustion is 1.401 g.
Note (add) and (total) were used subscript to "m"
Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe: PbO(s) NaCl(aq) H2O(l) --> Pb(OH)Cl(s) NaOH(aq) How many grams of PbO and how many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl
Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
[tex]= \frac{Mass}{Molar\ mass}[/tex]
[tex]= \frac{10.0 g}{259.65g / mol}[/tex]
= 0.0385 mol
Mass of PbO needed is
[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.