Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
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When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are: potassium hydrogen sulfate (aq) potassium hydroxide (aq) potassium sulfate (aq) water (l)\
Answer:
Explanation:
Answer:
1, 1, 1, 1
Explanation:
potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)
KHSO₄ + KOH ⟶ K₂SO₄ + H₂O
1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):
KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
2. Balance S:
We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.
1KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
3. Balance K:
We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + H₂O
4. Balance O
We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + 1H₂O
Every formula has a coefficient. The equation should be balanced.
5. Check that atoms balance:
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]
It checks.
The coefficients are 1, 1, 1, 1.
A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:Empirical formula ======== H₂O
Explanation:The empirical formula of a compound shows the whole number ratio for each atom in a compound.
To find empirical formula. we follow the below steps
The total mass of the compound here is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then assume 11.21 grams of hydrogen and 88.79grams of oxygen
Hydrogen Oxygen
1.composition by mass 11.21 88.79
molecular weight 1.007g/mol 15.990g/mol
2.Divide composition by mass 11.21/1.007 88.79/15.99
by each molecular weight to get 11.13 5.553
no of moles
3 Divide by the least number of moles
to get atomic ratio 11.13/5.553 5.553/5.553
2.004 1.00
4.Convert to whole numbers 2 1
Empirical formula ======== H₂O
Atomic mass is calculated by _____. subtracting protons from neutrons averaging the mass of isotopes adding protons and neutrons subtracting neutrons from protons
Answer:
Atomic mass is calculated by adding protons and neutrons.
Explanation:
Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).
We can craft an equation to show the relationship between these variables.
M - N = P, where M = Mass, N = Neutrons, and P = Protons
This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.
M = N + P
This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.
Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O
Answer:
Explanation:
Number of moles of propane:
=Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
= 44 * 6.8181
= 299.9964 grams
A number of moles of propane:
Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
=44 * 6.8181
= 299.9964 grams
What is carbon dioxide useful for?Carbon dioxide is used as a refrigerant, in fireplace extinguishers, for inflating lifestyles rafts and life jackets, blasting coal, foaming rubber and plastics, selling the increased vegetation in greenhouses, and immobilizing animals earlier than slaughter, and in carbonated liquids.
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For the following reaction, 3.76 grams of iron are mixed with excess oxygen gas . The reaction yields 4.29 grams of iron(II) oxide . iron ( s ) oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide
Answer:
4.84g of FeO is the theoretical yield
Explanation:
The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:
2Fe(s) + O₂(g) → 2FeO
Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.
To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:
Moles Fe (Molar mass: 55.845g/mol)
Using the molar mass of the compound we can convert grams to moles, thus:
3.76g Fe × (1mol / 55.845g) = 0.0673 moles of Fe
Moles and mass of FeO
As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.
Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:
0.0673 moles FeO × (71.844g / mol) =
4.84g of FeO is the theoretical yieldacid-catalyzed hydration of 1-methylcyclohexene gives two alcohols. The major product does not undergo oxidation, while the minor product will undergo oxidation. Explain
Answer:
Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is secondary alcohol.
Explanation:
Hello,
In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.
Best regards.
A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?
Answer:
74.3g of methanol were introduced into the vessel
Explanation:
In the equilibrium:
CH₃OH(g) ⇄ CO(g) + 2H₂(g)
Kc is defined as the ratio between concentrations in equilibrium of :
Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]
Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:
[CH₃OH] = ? - X
[CO] = X
[H₂] = 2X
Where ? is the initial concentration of methanol
As [H₂] = 2X = 0.426M; X = 0.213M
[CH₃OH] = ? - 0.213M
[CO] = 0.213M
[H₂] = 0.426M
Replacing in Kc to solve equilibrium concentration of methanol:
6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]
[CH₃OH] = 0.560
As:
[CH₃OH] = ? - 0.213M = 0.560M
? = 0.773M
0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:
3.00L * (0.773 mol / L) = 2.319 moles methanol.
Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:
2.319 moles * (32.04g / mol) =
74.3g of methanol were introduced into the vesselWhen 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
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The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.1 grams of 238U, but very little 235U. (a) What is the activity of the plate (in Ci)
Answer:
The correct answer is 1.68 × 10⁻⁵ Ci
Explanation:
The activity of the uranium is determined by using the formula,
R = 0.693 N/t1/2 -------------- (i)
The number of atoms is, N = nNA
Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,
N = (m/M) NA
The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,
N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³
The half-life of 238U from year to second is,
t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s
Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,
R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s
= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)
= 1.68 × 10⁻⁵ Ci
Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci
Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.
Answer:
[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]
Explanation:
You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.
1. Moles of B₄H₁₀
[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]
2. Molecules of B₄H₁₀
[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]
3. Atoms of H
[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]
Draw a structure for an alcohol that exhibits a molecular ion at M+ = 88 and that produces fragments at m/z = 73, m/z = 70 and m/z = 59.
Answer:
3-pentanol
Explanation:
In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].
Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at all). So, the structures only can be primary and secondary structures.
With this in mind, our options are:
-) 1-pentanol
-) 2-pentanol
-) 3-pentanol
Now we can analyze each structure:
-) 1-pentanol
The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 2-pentanol
On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 3-pentanol
In this structure, we have the "OH" bonded to carbon three. So, we can analyze each fragment:
-) m/z 59
This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.
-) m/z 71
This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.
-) m/z 73
This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.
See figure 2
I hope it helps!
Which ONE of these cations has the same number of unpaired electrons as Fe2+ ? A) Ni2+ B) Fe3+ C) Cr2+ D) Mn2+ E) Co2+
Answer:
Explanation:
Fe2+ Has 4 unpaired electrons.
By method of elimination;
Option A: Ni2+ has two unpaired electrons. so this option is wrong.
Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.
Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.
Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.
Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.
Classify each of the following fatty acids as saturated, monounsaturated, or polyunsaturated. Drag the appropriate items to their respective bins.
1. myristic acid
2. oleic acid
3. linoleic acid
4. palmitic acid
A. Saturated acids
B. Monounsaturated acids
C. Polyunsaturated acids
Answer:
A. Saturated acids - 1. myristic acid, 4. palmitic acid
B. Monounsaturated acids - 2. oleic acid
C. Polyunsaturated acids - 3. linoleic acid
Explanation:
Saturated fatty acids are the fats combined together with a single bond and no double or triple bond while unsaturated fatty acids are the fats having a double or triple bond between them.
Monosaturated acids have only one double or triple bond while polyunsaturated acids have more than one double or triple bond.
So, an appropriate match of given acids are:
A. Saturated acids - Myristic acid (CH3(CH2)12COOH) and Palmitic acid (CH3(CH2)14COOH) dont have any double or triple bonds.
B. Monounsaturated acids - Oleic acid (CH3(CH2)7CH=CH(CH 2)7COOH) have only one double bond.
C. Polyunsaturated acids - Linoleic acid (HOC–7CH=CH–CH 2–CH=CH–3H) has two double bonds.
Hence, the correct answer is:
A. Saturated acids - 1. myristic acid, 4. palmitic acid
B. Monounsaturated acids - 2. oleic acid
C. Polyunsaturated acids - 3. linoleic acid
How many cups are in five gallons?
Answer:
In 5 US liquid gallons, there are 80 cups.
Explanation:
To get from gallons to cups, just multiply the amount of gallons you have by 16.
What are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
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The two common properties of all solids are fixed _____ and _____.
Answer:
shape
volume
Hope this helps! （づ￣3￣）づ╭❤～
Explanation:
1. What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0x10-14), with a concentration of
hydroxide ions of 2.21x10-6 M? A. 3.1x10-6 M
B. 4.52 X10-9 M
C. 2.8x10-8 M
D. 1.6x10-9 M
Answer:
B. 4.52 X10-9 M
Explanation:
Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:
[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]
If we write the Keq expression for this reaction we will have:
[tex]Keq=[H^+][OH^-][/tex]
Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":
[tex]Kw=[H^+][OH^-][/tex]
From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:
[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]
[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]
I hope it helps!
Answer:
B. 4.52 * 10^-9M
Explanation:
did the test
Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
Determine whether each chemical substance would remain the same color or turn pink in the presence of phenolphthalein.
Answer:
See the answer below
Explanation:
The complete question can be seen in the attached image.
Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.
Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;
1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.
2. Battery acid with a pH of 1 is acidic, it will remain colorless.
3. Lime juice with a pH of 2 is acidic, it will remain colorless.
4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.
5. Seawater with a pH of 8.5 is basic, it will turn pink.
6. Tap water with a pH of 7 is neutral, it will remain colorless
Phenolphthalein is a chemical compound with the formula[tex]C_{20}H_{14}O_4[/tex]. Phenolphthalein is often used as an indicator in acid-base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions
Phenolphthalein works as in:-
Colorless in acidPink in baseAccording to the question, There are 5 solutions having different ph and the indication only turns basic solution to pink
The indicator only turn the basic solution pink and these solutions are as follows,
AmmoniaSea waterTap water.Hence, these are the answer.
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What is silica gel commonly used for? A. Absorbing moisture to protect goods from damage. B. As insulation in buildings. C. As a lacquer on wood to make it water-resistant. D. A soft, flexible padding, such as on pen grips or mouse pads.
Answer:
A
Explanation:
Absorbing moisture to protect goods from damage. Hence, option A is correct.
What is silica gel?Silica gel is a desiccant, or drying agent, that manufacturers often place in little packets to keep moisture from damaging certain food and commercial products.
Silica Gel is a good drying agent for preventing corrosion, contamination, spoilage, and mould growth in many commodities and products due to its physical properties.
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Gaseous indium dihydride is formed from the elements at elevated temperature:
In(g)+H2(g)⇌InH2(g),Kp=1.48 at 973 K
The partial pressures measured in a reaction vessel are
PIn =0.0540atm
PH2= 0.0250atm
PInH2 =0.0780atm
Calculate Qp and give equal partial pressure for In, H2, and InH2.
Answer:
The reaction given is:
In (g) + H₂ (g) ⇔ InH₂ (g), the Kp is 1.48 at 973 K.
The partial pressures measured in the reaction vessel is Partial pressure of In is 0.0540 atm, partial pressure of H₂ is 0.0250 atm, and the partial pressure of InH₂ is 0.0780 atm. By using the table given in the attachment below, the value of PInH₂ is (0.078-x), PIn is (0.054 + x), and the value of PH2 is (0.025 + x).
Kp = PInH₂/PIn × PH₂ = (0.078 - x) / (0.054 +x) (0.025 + x)
1.48 = (0.078 - x) / (0.054 +x) (0.025 + x)
x = 0.06689
Now the partial pressures of In, H₂ and InH₂ will be,
PH₂ = 0.025 + x = 0.025 + 0.0668 = 0.0918 atm
PIn = 0.054 + 0.0668 = 0.1208 atm
PInH₂ = 0.078 - 0.0668 = 0.0112 atm
Now the Qp or the reaction quotient will be,
Qp = (0.078) / (0.054) (0.025) = 57.78.
what js the percent yield of lithium hydroxide from a reaction of 7.40 g of lithium with 10.2 g of water? the actual yield was measured to be12.1 g
Answer:
89%.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2Li + 2H2O —> 2LiOH + H2
Next, we shall determine the masses of Li and H2O that reacted and the mass of LiOH produced from the balanced equation.
This is illustrated below:
Molar mass of Li = 7 g/mol
Mass of Li from the balanced equation = 2 x 7 = 14 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol
Mass of LiOH from the balanced equation = 2 x 24 = 48 g
Summary:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O to produce 48 g of LiOH.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O.
Therefore, 7.4 g of Li will react with = (7.4 x 36)/14 = 19.03 g of H2O.
From the calculation made above, we can see that it will take a higher amount i.e 19.03 g than what was given i.e 10.2 g of H2O to react completely with 7.4 g of Li.
Therefore, H2O is the limiting reactant and Li is the excess reactant.
Next, we shall determine the theoretical yield of LiOH.
In this case we shall use the limiting reactant.
The limiting reactant is H2O and the theoretical yield of LiOH can be obtained as follow:
From the balanced equation above,
36 g of H2O reacted to produce 48 g of LiOH.
Therefore, 10.2 g of H2O will react to produce = (10.2 x 48)/36 = 13.6 g of LiOH.
Therefore, the theoretical yield of LiOH is 13.6 g
Finally, we shall determine the percentage yield of LiOH. This can be obtained as follow:
Actual yield = 12.1 g
Theoretical yield = 13.6 g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 12.1/ 13.6 x 100
Percentage yield = 89%
Therefore, the percentage yield LiOH is 89%.
What is the electron configuration for the transition metal ion in each of the following compounds?
[Ni(H2O)6]Br2
[Cr(H2O)4(NO2)2]I
Answer:
1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled
2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled
g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–CH3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Answer:
C) formaldehyde, H2C=O.
Explanation:
Hello,
In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).
Best regards.
A chamber contains equal molar amounts of He, Ne, Ar, and Kr. If the total chamber pressure is 1 atm, then the partial pressure (in atm) of Kr is:
Answer:
The correct answer is 0.25 atm.
Explanation:
As mentioned in the given question that the chamber comprises equal molar concentrations of He, Ne, Ar and Kr gas. So, let us assume that the moles of all the gases will be x then the total number of moles will be 4x.
The formula for calculating mole fraction is,
Mole fraction = mole of the substance/total moles
The mole fraction of Kr = x/4x = 0.25
The total pressure given in the chamber is 1 atm. Therefore, the partial pressure will be,
Partial pressure = mole fraction * Total pressure
Partial pressure = 0.25 * 1 = 0.25 atm.
The partial pressure of Kr is 0.25 atm.
We are told that there is an equal molar amounts of He, Ne, Ar, and Kr. If we decide to say, let the molar amount of each gas be x, the total number of moles of all the gases will now be; x + x+ x + x = 4x
Let us recall that partial pressure of a gas can be obtained form the formula;
Partial pressure = mole fraction × total pressure
Mole fraction of Kr = x/4x = 1/4
Total pressure = 1 atm
Partial pressure of Kr = 1/4 × 1 = 0.25 atm
Learn more: https://brainly.com/question/2510654
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Answer:
14, 508J/K
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
m = mass of substance (g) = 0.1184g
1 mole of Mg - 24g
n moles - 0.1184g
n = 0.0049 moles.
Also, q = m × c × ΔT
Heat Capacity, C of MgCl2 = 71.09 J/(mol K)
∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)
= 14, 508 J/K/kg
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg
= 1,7117.7472 J °C-1 g-1
∴ ΔHrxn = q/n
=1,7117.7472 ÷ 0.1184
= 14, 508J/K
An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N2(g) + 3 H2(g)<=> 2NH3(g) is 0.29.
What is the concentration of NH3?
Answer:
5.94×10¯³
Explanation:
The following data were obtained from the question:
Concentration of N2, [N2] = 0.036 M
Concentration of H2, [H2] = 0.15 M
Equilibrium constant (Kc) = 0.29 M
Concentration of NH3, [NH3] =.…?
The equation for the reaction is given below:
N2(g) + 3H2(g) <=> 2NH3(g)
Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.
This is illustrated below:
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant for the above equation is given below:
Kc = [NH3]² / [N2] [H2]³
Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:
Kc = [NH3]² / [N2] [H2]³
0.29 = [NH3]²/ 0.036 × 0.15³
Cross multiply
[NH3]² = 0.29 × 0.036 × 0.15³
[NH3]² = 3.5235×10¯⁵
Take the square root of both side
[NH3] = √(3.5235×10¯⁵)
[NH3] = 5.94×10¯³
Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.
In the experiment students will create solutions with different ratios of ethanol and water. What is the mole fraction of ethanol when 10.00 mL of pure ethanol is combined with 2.00 mL of water
Answer:
[tex]x_{et}=0.6068[/tex]
Explanation:
Hello,
In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:
[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]
Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)
[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]
Therefore, the mole fraction turns out:
[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]
Best regards.
A solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium? Express the pH numerically.
Answer:
[tex]pH=2.28[/tex]
Explanation:
Hello,
In this case, for the acid dissociation of formic acid (HCOOH) we have:
[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]
Whose equilibrium expression is:
[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
That in terms of the reaction extent is:
[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]
Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:
[tex]x=0.00528M[/tex]
[tex][H^+]=0.00528M[/tex]
Then, as the pH is computed as:
[tex]pH=-log([H^+])[/tex]
The pH turns out:
[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]
Regards.
