Answer:
b
Explanation:
soil mixed with water sliding off a cliff
The acid dissociation constant Ka equals 1.26 × 10–2 for HSO4– (acid 1) and is 5.6 × 10–10for NH4+(acid 2). Predict the net direction of the following reaction: HSO4–(aq) + NH3(aq) SO42–(aq) + NH4+(aq)
Answer:
As K >>> 1, the reaction will shift to the products
Explanation:
To know the direction of any reaction you must calculate the equilibrium constant, K. If K is < 1, the reaction will shift to the reactants and if k > 1 the reaction will shift to the products.
With the reactions:
HSO₄⁻ ⇄ SO₄²⁻ + H⁺ Ka = 1.26x10⁻²
And:
NH₄⁺ ⇄ NH₃⁺ + H⁺ Ka = 5.6x10⁻¹⁰
The inverse reaction:
NH₃⁺ + H⁺ ⇄ NH₄⁺ 1/Ka = 1.8x10⁹
The sum of the reactions:
HSO₄⁻ + NH₃⁺ + H⁺ ⇄ NH₄⁺ + SO₄²⁻ + H⁺ K = 1.26x10⁻² ₓ 1.8x10⁹ = 2.3x10⁷
As K >>> 1, the reaction will shift to the productsWhat volume of 0.200 M HCl will react with 0.2500 g of CaCO3? CaCO3 (s) + 2 HCl (aq) ➔ CaCl2(aq) + H2O (l) + CO3 (g).
Answer:
0.02498 L
Explanation:
CaCO₃ + 2 HCl ➔ CaCl₂ + H₂O + CO₂
You have to use stoichiometry. According to the chemical equation, for every one mole of CaCO₃, two moles of HCl is needed for the reaction to occur. Before you can use this relation you need to convert grams of CaCO₃ to moles. To convert, you need to use the molar mass.
Molar mass of CaCO₃ = 100.086 g/mol
(0.2500 g)/(100.086 g/mol) = 0.002498 mol CaCO₃
Now using the relation of 1 mol of CaCO₃ for every 2 mol of HCl, convert moles of CaCO₃ to moles of HCl.
(0.002498 mol CaCO₃) (2 mol HCl)/(1 mol CaCO₃) = 0.004996 mol HCl
Since the molarity of the solution of HCl is 0.200 M (mol/L), you have to divide the amount of moles needed by the molarity of the solution.
(0.004996 mol HCl)/(0.200 M) = 0.02498 L
You will need 0.02498 L to react with 0.2500 g of CaCO₃.
What is the half-life for the first order decay of 14C according to the reaction, 146C — 147N +e- ?
The rate constant for the decay is 1.21 x10-4 year-1
Answer:
5727 years or 5730 (rounded to match 3 sig figs) whichever one your teacher prefers
Explanation:
First Order decay has a half life formula of Half Life = Ln (2) / k = 0.693/K
Half-life = 0.693/k = 0.693/1.21 x10-4 = 5727 years or 5730 (rounded to match 3 sig figs)
This should be correct because if you google the half-life of 14 C it is ~ 5700 years
A student has a sample of CaSO4 hydrate and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration, and after subsequent heatings, the student finds the anhydrous compound has a constant mass of 0.3750 grams. Find the formula of the hydrate.(2 points)
Answer:
CaSO4•3H2O.
Explanation:
Let the compound be CaSO4.xH2O.
The following data were obtained from the question:
Mass of hydrated compound (CaSO4.xH2O) = 0.4813g
Mass of anhydrous compound (CaSO4) = 0.3750g
Next we shall determine the mass of the water is the hydrated compound.
This is illustrated below:
Mass of water = mass of hydrated – mass of anhydrous.
Mass of water = 0.4813 – 0.3750
Mass of water = 0.1063g
Next, we shall determine the number of mole of the anhydrous compound and the number of mole of the water present in the compound. This is illustrated below:
Molar mass of anhydrous CaSO4 = 63.5 + 32 + (16x4) = 159.5g/mol
Mass of anhydrous CaSO4 = 0.3750g
Mole of anhydrous CaSO4 =...?
Mole = mass /Molar mass
Mole of anhydrous CaSO4 = 0.3750/159.5 = 2.35×10¯³ mole
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O = 0.1063g
Number of mole of H2O =.?
Mole = mass /Molar mass
Mole of H2O = 0.1063/18 = 5.91×10¯³ mole
Next we shall determine the ratio of number of mole of anhydrous CaSO4 to that of H2O. This is illustrated below:
Mole of anhydrous CaSO4 = 2.35×10¯³ mole
Mole of H2O = 5.91×10¯³ mole
Ratio of anhydrous CaSO4 to H2O =>
CaSO4 : H2O => 2.35×10¯³ /5.91×10¯³
CaSO4 : H2O => 1 : 3
Therefore, for 1 mole of the anhydrous CaSO4, there are 3 moles of H2O.
Therefore, the formula for the hydrate compound CaSO4.xH2O => CaSO4•3H2O.
why we used petrol for vehicles not water?
Answer:
Water isn't combustible. There is nothing you can add to it other than gasoline that will make it even remotely combustible. Now, through electrolysis, it can be broken down into hydrogen and oxygen, which could be burned in an internal combustion engine.
because the water in itself does not produce energy
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 name of CaCl2 7H2o
Answer:
calcium chloride dihydrate
What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)
Answer:
[tex]m=59.3gCO_2[/tex]
Explanation:
Hello,
In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:
[tex]M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol[/tex]
It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:
[tex]mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol[/tex]
Finally, the mass by using the molar mass:
[tex]m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2[/tex]
Best regards.
Answer:
59.3g
Explanation:
Got it right in Gizmo
Which element has the greatest electronegativity . Ci , c , b
Answer:
CI
Explanation:
it is closer to fluorine
HELP ASAP if you're good at gen chem I /AP chem: A student wishes to prepare a solution with a final concentration of Na+ = 0.50 M and a final concentration of HCO3– = 0.10 M by taking some NaOH and some trona (Na2CO3•NaHCO3•2H2O) and diluting with water to a final volume of 1.00 L. How much NaOH and trona are required? the answer is 0.05 mol NaOH and 0.15 mol trona, but I just don't understand how in the world to get there!?
Answer:
Moles of trona are 0.15moles
Moles of NaOH are 0.05mol
Explanation:
In the problem, you have 2 solutes, NaOH and trona.
Based in its formula, in solution, 1 mole of NaOH will produce 1 mole of Na⁺ and 1 mole of trona will produce 3 moles of Na⁺ and 1 mole of HCO₃⁻.
Also, you must know HCO₃⁻ will react with NaOH producing CO₃²⁻.
You want 1L of 0.50M Na⁺ = 0.50 moles of Na⁺ and 1L of HCO₃⁻ = 0.10M.
Now, in equations, moles of Na⁺ will be:
Moles Na⁺ = 0.50moles = X + 3Y (1)
Where X are moles of NaOH and Y moles of trona.
And, moles of HCO₃⁻ will be:
Moles HCO₃⁻ = 0.10moles = Y - X (2)
The sum of (1) + (2):
0.60mol = 4Y
0.15mol = Y
Moles of trona are 0.15molesAnd moles of NaOH are (Replacing in (1) or (2) you will obtain the same, I will replace in (1)) :
0.50moles = X + 3Y
0.50moles = X + 3×0.15mol
0.50moles = X + 0.45mol
0.05 moles = X
Moles of NaOH are 0.05molwhen 1.00g of magnesium is reacted with excess hydrochloric acid, what volume of hydrogen gas will be produced at standard temperature and pressure? statndard temperature and pressure is defined as 0 C and 1 atm. At STP, 1 mole of gas occupies 22.4L.
Answer:
0.93L
Explanation:
Hello,
To solve this question, we need to first of all write down the equation of reaction.
Equation of reaction,
2Mg + 4HCl → 2MgCl₂ + 2H₂
From the equation of reaction, it shows that 2 moles of Mg produces 2 moles of H₂.
Number of moles = mass / molar mass
Mass = number of moles × molar mass
Molar mass of Mg = 24g/mol
Mass = 2 × 24 = 48g
Therefore, 48g of Mg produces (2 × 22.4)L of H₂
48g of Mg = 44.8L of H₂
1g of Mg = xL of H₂
x = (1 × 44.8) / 48
x = 0.93L
1g of Mg will produce 0.93L of H₂
How many milliliters of a 0.150 M NaOH solution must be added to a 25.0-mL sample of 0.150 M butanoic acid, CH3(CH2)2CO2H, in order to create a buffer with a pH of 4.82? Ka of butanoic acid = 1.5 x 10-5
Answer:
12.5 mL of NaOH
Explanation:
0.150 M NaOH
CH3-CH2-CH2-COOH 0.150M
This is the data given. The reaction of both reactants make a neutralization and a salt
CH3-CH2-CH2-COOH + NaOH → NaCH3-CH2-CH2-COO + H₂O
As both compounds produce a buffer, we can apply the Henderson Hasselbach equation:
pH = pKa + log ( mmol NaOH / Initial mmol of acid - mmol NaOH)
Final mmol of acid = mmol of NaOH
Final mmol of acid = Initial mmol of acid - mmol of NaOH we added.
Let's find the data to replace in the formula:
pKa butanoic acid = - log Ka → - log 1.5×10⁻⁵ = 4.82
Initial mmol of acid = M . mL → 25 mL . 0.150M → 3.75 mmol
Notice that M = mmol/mL → mmol = M . mL.
We replace → 4.82 = 4.82 + log mmol NaOH / ( 3.75 mmol - mmol NaOH)
As pH = pKa, notice that log = 1 so:
mmol NaOH / ( 3.75 mmol - mmol NaOH) = 1
mmol NaOH = 3.75 mmol - mmol NaOH
2mmol NaOH = 3.75 → mmol NaOH = 3.75 / 2 = 1.875
M = mmol /mL → mL = mmol / M → 1.875 mmol / 0.150M = 12.5 mL
Calculate the empirical formula for each of the following substances. (Express answer as a chemical formula) 1) 2.90 g of Ag and 0.125 g of N 2) 2.22 g of Na and 1.55 g of S 3) 2.11 g of Na, 0.0900 g of H, 2.94 g of S, and 5.86 g of O 4) 1.84 g of K, 0.657 g of N, and 2.25 g of O
Answer:
1) Ag3N
2)Na2S
3)NaHSO4
4) KNO3
Explanation:
We divide each mass by the element's relative atomic mass
1) 2.90/108-Ag, 0.125/14-N
0.027-Ag, 0.0089-N
Divide by the lowest ratio
0.027/0.0089-Ag, 0.0089/0.0089 N
3-Ag, 1-N
Empirical formula- Ag3N
2)2.22/23-Na, 1.55/32-S
0.097-Na, 0.048-S
Divide by the lowest ratio
0.097/0.048-Na, 0.048/0.048-S
2-Na, 1-S
Empirical formula- Na2S
3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O
0.09-Na, 0.09-H, 0.09-S,0.366-O
Divide by the lowest ratio
0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O
1-Na, 1-H, 1-S, 4-O
Empirical formula- NaHSO4
4)1.84/39, 0.657/14-N, 2.25/16-O
0.047-K, 0.047-N, 0.14-O
Divide through by the lowest ratio
0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O
1-K, 1-N, O-3
Empirical formula- KNO3
Why at night, under the mercury or sodium vapor lights in a mall parking lot, do cars seem to be peculiar colors?
Answer:
Here's what I find
Explanation:
Sodium vapour produces mostly yellow light.
Mercury vapour produces mostly blue light.
A white car reflects all colours back to our eyes.
A coloured car, say green, absorbs all the other colours and reflects the green.
Under a sodium light, a white or a yellow car will appear yellow. A red car will be nearly black because there is no red light to reflect.
Under a mercury light, a white or a blue car will appear blue. A yellow car will be quite dark, but there will be a yellowish tinge because there is some yellow in the Hg spectrum for it to reflect.
The picture below shows a red car and a black car under sodium light. Can you tell which is which?
Draw the structure of the organic product(s) of the Grignard reaction between dimethyl carbonate (CH3OCO2CH3) and excess phenylmagnesium bromide, followed by aqueous workup. You do not have to consider stereochemistry. If a compound is formed more than once, add another sketcher and draw it again. Alternatively, you may use the square brackets tool to add stoichiometries greater than one. Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner. Separate multiple products using the + sign from the dropdown menu.
Answer:
dimethoxy(phenyl)methanol
Explanation:
For this question, we have to remember the mechanism of the Grignard reaction. In this case, phenylmagnesium bromide is our nucleophile, a carbo-anion is produced (step 1). Then this carbo-anion can attack the carbonyl group in the dimethyl carbonate, the double bond is delocalized into the oxygen producing a negative charge (step 2). Finally, with the addition of the hydronium ion ([tex]H_3O^+[/tex]), the anion can be protonated to produce the alcohol (dimethoxy(phenyl)methanol) (step 3).
See figure 1
I hope it helps!
At a temperature of 393 K, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at a temperature of 478 K? (Assume constant volume)
Answer:
1.30atm
Explanation:
P1/T1 = P2/T2
1.07/393 = P2/478
Answer: the first one is correct
Explanation:
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A scientist isolates 2.391g of a gas. The sample occupies a volume of 562.3mL at 94.8 degrees celsius and 108.2 kPa. Calculater the molae mass of the gas. please show all your steps
Answer:
molar mass = 129.55g/mol
Explanation:
using ideal gas equation
PV = nRT
P = 108.2kpa = 108200pa = 1.07atm
v = 562.6mL = 562.6 x 10^-³L
R = 0.082057 L atm mol-1K-1,
mass = 2.391g
T = 94.8°C = 367.8K
molar mass = unknown
n = mass/molar masspv= (mass/molar) RT1.07 x 562.6 x 10^-³ = (2.391/molar mass) x 0.08206 x 367.8(0.602) = (2.391/molar mass) x 30.1820.602/30.182= (2.391/molar mass)0.02= (2.391/molar mass)molar mass = (2.391/0.02)molar mass = 129.55g/molThe chemical reaction that occurs in a gas grill is the combustion of propane, C3H8. Write a balanced equation for this reaction. (Include states of matter.)
Answer:
[tex]\Large \boxed{\mathrm{C_3 H_8 \ (g)+5O_2 \ (g) \Rightarrow 3CO_2 \ (g)+4 H_2 O \ (l)}}[/tex]
[tex]\rule[225]{225}{2}[/tex]
Explanation:
[tex]\sf C_3 H_8 +O_2 \Rightarrow CO_2 + H_2 O[/tex]
Balancing Carbon atoms on the right side,
[tex]\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 + H_2 O[/tex]
Balancing Hydrogen atoms on the right side,
[tex]\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 +4 H_2 O[/tex]
Balancing Oxygen atoms on the left side,
[tex]\sf C_3 H_8 +5O_2 \Rightarrow 3CO_2 +4 H_2 O[/tex]
[tex]\rule[225]{225}{2}[/tex]
The combustion reaction of propane is :C₃H₈ [tex]_(g)[/tex]+5 O₂ [tex]_(g)[/tex][tex]\rightarrow[/tex]3 CO₂[tex]_(g)[/tex] + 4 H₂O[tex]_(l)[/tex].
What is combustion reaction?Combustion is defined as a high temperature exothermic reaction between a fuel which acts as a reductant and an oxidant which is usually atmospheric oxygen.It does not result in fire as the flame is visible only when substance which undergoes combustion vaporizes.
Activation energy must be overcome so that combustion is initiated.Solid fuels such as wood and coal initially undergo endothermal pyrolysis which results in gaseous fuels. It is widely used in industry and the application is based on concepts of chemistry, physics and mechanics.
It is a complex chemical process involving many steps which depend on properties of combustible material. It is a type of redox reaction.
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Balance the chemical equation
Fe2O3 (s) + CO (g) 2 Fe(s) + CO2 (g)
Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
[tex]Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]
Explanation:
[tex]Fe_2O_3+CO\Rightarrow \:2Fe+CO_2\\\\Fe_2O_3+3CO\Rightarrow \:2Fe+3CO_2[/tex]
Best Regards!
Which substance is a base? a. HCOOH b. RbOH c. H2CO3 d. NaNO3
Answer:
rboh
Explanation:copied from previous
Among the following , RbOH is a base as it has hydroxyl ions which are generated on dissociation.
What is a base?
According to the Arrhenius concept, base is defined as a substance which yields hydroxyl ions on dissociation.These ions react with the hydrogen ions of acids to produce salt in an acid-base reaction.
Bases have a pH higher than seven as they yield hydroxyl ions on dissociation.They are soapy in touch and have a bitter taste.According to the Lowry-Bronsted concept, base is defined as a substance which accepts protons .Base react violently with acids to produce salts .Aqueous solutions of bases can be used to conduct electricity .They can also be used as indicators in acid-base titrations.
They are used in the manufacture of soaps,paper, bleaching powder.Calcium hydroxide ,a base is used to clean sulfur dioxide gas while magnesium hydroxide can be used as an antacid to cure acidity.
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chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12×10−12 . What is the solubility (in mol/L) of silver chromate in 1.00 M potassium chromate aqueous solution?
Answer:
[tex]6.5X10^-^5~\frac{mol}{L}[/tex]
Explanation:
For this question, we have to start with the ionization equation for [tex]Ag_2CrO_4_(_s_)[/tex], so:
[tex]Ag_2CrO_4_(_s_)~<->~2Ag^+~_(_a_q_)~+~CrO_4^-^2_(_a_q_)[/tex]
With this in mind we can write the Ksp expression:
[tex]Kps~=~[Ag^+]^2[CrO_4^-^2][/tex]
Additionally, for every mole of [tex]CrO_4^-^2[/tex] formed, 2 moles of [tex]Ag^+[/tex] are formed. We can use "X" for the unknown concentration of each ion, so:
[tex][CrO_4^-^2]~=~X[/tex] and [tex][Ag^+]~=~2X[/tex]
Now, we can plug the values into the Ksp expression:
[tex]1.12x10^-^1^2~=~(2X)^2(X)[/tex]
Now we can solve for "X" :
[tex]1.12x10^-^1^2~=~4X^3[/tex]
[tex]X^3=\frac{1.12X10^-^1^2~}{4}[/tex]
[tex]X=(\frac{1.12X10^-^1^2~}{4})^(^1^/^3^)[/tex]
[tex]X=6.5X10^-^5~\frac{mol}{L}[/tex]
I hope it helps!
2.7g of aluminum metal is placed into 200g of
solution hydrochloric acid to form aluminum
chloride and hydrogen gas.
a. Write the chemical equation.
b. Find the mass of the obtained aluminum
chloride.
C. Find the percent by mass of the acidic solution.
d. Find the volume of the obtained hydrogen gas
at STP.
Answer:
a.
2Al + 6HCl--> 2AlCl3 + 3H2
b.
take the atomic mass of Al =27.0, Cl = 35.5, and H= 1.0,
no. of moles = .mass / molar mass
no. of moles of Al placed = 2.7 / 27.0
=0.1 mol
Assuming Al is limiting (because the concentration nor volume of HCl solution is given).
From the equation, the mole ratio of Al : AlCl3 = 2:2 = 1: 1,
meaning for each 1 mole of Al reacted, it gives out 1 mole of AlCl3. This also means the no. of moles of Al reacted equals to the no. of moles of AlCl3 produced, which is 0.1 mole.
mass = no. of moles x molar mass
hence,
mass of AlCl3 produced = 0.1 / (27 + 35.5x3)
= 0.00075 g
c. since this question is unclear, and the concentration nor volume of HCl solution is given, part c cannot be answered. Please let me know if you have any more info regarding it! thanks :)
d. From the equation, the mole ratio of Al:H2 = 2:3, meaning every 2 moles of Al reacted, gives out 3 moles of hydrogen.
using this ratio, no. of moles of H2 produced will = 0.1 x 3 / 2
=0.15 mol
At STP, the molar volume of gas is 22.4 dm3 mol-1,
volume = no. of moles x molar volume
hence volume of H2 obtained = 0.15 x 22.4
= 3.36 dm3
How many valence electrons should be used to draw a Lewis structure of methane CH4
Answer:
8 valence electrons
Explanation:
There are 4 valence electrons in carbon because it is in group 4A of the periodic table. There is 1 valence electron per hydrogen based on its position on the periodic table, but CH4 has 4 hydrogens in it. So 1 times 4 is 4. Add the 4 valence electrons from carbon with the 4 electrons from H4 to get 8.
Answer:
the answer is: 8
Explanation:
i took the test :D
A chemistry student weighs out of hypochlorous acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
Answer:
Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.
A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.
Explanation:
1 mole HClO = 74.44g
0.0941g = [tex]\frac{0.0941}{74.44}[/tex] = 0.00126 moles
Concentration = no. of moles/volume in L
Hence, Concentration of HClO = 0.00126/ 0.250L
= 0.005M.
C1V1 =C2V2
0.005 × 250 mL = 0.2 × V2
Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.
structure and correct name for 5-octyne
Answer:
Explanation:
(R)-5-octyne-4-ol
C8H140
Correct name for 5-octyne is 3-octane.
Structure is attached below.
3-Octane:Octane is a hydrocarbon and an alkane with the chemical formula C₈H₁₈, and the condensed structural formula CH₃(CH₂)₆CH₃. Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.
The structure for 5-octyne is given below.
Correct name for it is 3-octane.
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correctly complete the following statement: The side chains ('-R groups') of amino acids in a beta-sheet: A. alternate pointing "up" and "down" for each successive amino acid in the primary sequence. B. point in one direction in a parallel beta sheet and the opposite direction in an antiparallel beta-sheet. C. always point toward the direction of water. D. point in random directions. E. all point in opposite directions
Answer:
A.
Explanation:
The β-sheet is a twisted pattern in which the protein strands are laterally linked through hydrogen bonds. These β-sheet motifs are segments of 3 to 10 amino acids that are specially configured to form beta antiparallel strands. In β-sheet regions, the hydrogen bonds are localized among carbonyl and amino groups of the polypeptide backbone, and side chains (i.e., the R groups) are extended up and down in the plane of the β-sheet.
1. Draw a wedge/dash structure for trans-1,2-dimethylcyclohexane.
2. Draw the chair structure of the compound.
3. Draw the ring-flipped structure of the compound.
4. Identify which of these structures is the most stable or if both are equal in energy.
Answer:
The structure with the ring flipped is the most stable
Explanation:
We have the trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.
The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.
In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.
On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.
Hope this helps
The composition of planetary atmospheres is determined in part by the speeds of the molecules of the constituent gases because the faster-moving molecules can reach escape velocity and leave the planet. Calculate the mean speed of CH4 molecules at 77K and 1500K.
Answer:
At T = 77k, V = 2.69m/a
At T = 1500K, V = 11.86m/s
Explanation:
Hello,
We're required to find the root mean speed (RMS) of a gas and to do that, we'll have to relate the energy of a gas and its kinetic energy
Ek = 3/2 RT
Ek = ½mv²
½mv² = 3/2 RT
Solve for V
MV² = 3RT
V² = 3RT / M
V = √(3RT/M)
V = velocity or speed of the gas
R = ideal gas constant
T = temperature of the gas
M = molarmass mass of the gas
Molar mass of CH4 = 12 + 4 = 16g/mol
At T = 77k
V = √[((3/2) × 77) / 16]
V = √7.216
V = 2.69m/s
At T = 1500K
V = √[((3/2) × 1500) / 16]
V = √140.625
V = 11.86m/s
Resources Use the exothermic and endothermic interactive to classify the solution process for each solute. Exothermic solution process Endothermic solution process
KOH CaCl, NaCT NaOH NaNO, NH NO,
Answer:
Exothermic interractive are the following: NaOH, KOH, CaCl₂
Endothermic interactive are the following: NaCl, NH₄NO₃, NaNO₃
Explanation:
NaOH, KOH, and CaCl2 are exothermic reactants. NaCl, NH4NO3, and NaNO3 are endothermic interacting substances.
Endothermic reactions: what are they?Chemical processes that can release or absorb energy are referred to as endothermic. In endothermic reactions, more energy is used when bonds in the reactants are broken than is released when new bonds are formed in the products.
How do endothermic processes take place?When the temperature of the an isolated system drops while the surroundings of the a non-isolated system warm up, this is known as an endothermic response. The heat of reaction is generally positive in endothermic processes (qrxn>0).
To know more about endothermic reactions visit:
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A 10.0 L balloon contains helium gas at a pressure of 655 mmHg. What is the new pressure, in mm Hg, of the helium gas if the volume is 20.0 L and if there is no change in temperature or quantity? 6550 mmHg 1000 mm Hg 1310 mmHg 300 mm Hg
Answer:
327.5 mmHg
Explanation:
By Boyle's law, P1V1=P2V2
10 L * 655 mmHg = 20 L * x
solve for x
x= (10*655)/20
x=327.5 mmHg
