Answer:
pH of the solution is 2.0
Explanation:
The lactic acid is a weak acid that is in equilibrium with water as follows:
Lactic acid + H2O ⇄ Lactate + H₃O⁺
And Ka for lactic acid: 1.38x10⁻⁴
Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]
Initial concentration of lactic acid is (MW: 112.06g/mol):
80g * (1mol / 112.06g) / 1L = 0.714M
The equilibrium concentration of the species in the equilibrium are:
[Lactate] = X
[H₃O⁺] = X
[Lactic acid] = 0.714-X
Replacing in Ka expression:
1.38x10⁻⁴ = [X] [X] / [0.714-X]
9.8532x10⁻⁵ - 1.38x10⁻⁴X = X²
9.8532x10⁻⁵ - 1.38x10⁻⁴X - X² = 0
Solving for X:
X = -1.0x10⁻². False solution, there is no negative concentrations
X = 9.86x10⁻³M. Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 9.86x10⁻³M
and pH = -log [H₃O⁺] = -log 9.86x10⁻³M
pH = 2.0
pH of the solution is 2.0For the following reaction, predict whether the equilibrium lies predominantly to the left or to the right. Explain.
NH4+(aq) + Br-(aq) - NH3(aq) + HBr(aq)
If Ka
(NH4+) = 5.6 x 10-10
Answer:
Lies predominantly to the left.
Explanation:
In the reaction:
NH4⁺(aq) + Br-(aq) ⇄ NH3(aq) + HBr(aq)
Conjugate acid + Ion ⇄ weak base + strong acid
HBr is a strong acid whereas NH3/NH4⁺ are the weak base and its conjugate base. A strong acid as HBr dissociates completely in solution as H⁺ and Br⁻. That means in solution you will never have HBr without dissociation doing the reaction:
lies predominantly to the left.Define the following terms - you may need to consult your lecture text or other suitable resource:
a. monomer,
b. repeating unit,
c. condensation polymerization,
d. cross-linked polymer
Answer:
a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.
b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.
c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.
d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers. The cross-links leads to an alteration in the physical properties of the polymer.
The definition of following terms are :
a) Monomers:
The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.
The polymer shaped can too be in a three dimensional arrange.
b) Repeating unit:
A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.
A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.
c) Condensation polymerization:
This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.
d) Cross-linked polymer:
This can be a polymer shaped from a sort of holding of particles.
The cross-links leads to an modification within the physical properties.
DefinitionsDefinition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.
Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.
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What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH
Answer:
0.0630
Explanation:
The molar mass of urea = 60 g/mol
we all know that:
[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
Then; the number of moles of urea
= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]
= 0.0667 mol
Similarly; the number of moles of methanol
= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]
= 0.9988 mol
The total number of moles = (0.0667 + 0.9988) mol
= 1.0655 mol
Finally,the mole fraction of urea [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]
[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]
= 0.0630
Which of the following provides a characteristic of
MgO(s) with a correct explanation?
Choose 1 answer:
А
It is hard because its ions are held together by strong
electrostatic attractions.
B
It is malleable because its atoms can easily move past
one another without disrupting the bonding.
It is a poor conductor of electricity because its
electrons are tightly held within covalent bonds and
lone pairs.
It has a high melting point because its molecules
interact through strong intermolecular forces.
Answer:
А It is hard because its ions are held together by strong electrostatic attractions.
B It is malleable because its atoms can easily move past one another without disrupting the bonding.
Explanation:
These are correct explanations of the properties of magnesium.
C is wrong. Mg is a good conductor of electricity and it has metallic bonds.
D is wrong. Mg has no molecules. It has no intermolecular forces.
Complete the unit conversion by entering the correct numbers
A=
B=
C=
Answer: A=1, B=3, C=12
Explanation:
For this problem, you will need to know your unit conversions. There are 3 ft in 1 yard. Knowing this, we can find A, B, C.
For A and B, we know that we want to cancel out ft so the answer can be in yards. To do so, we need to put B=3 and A=1.
Now that we know the unit conversion, we can directly solve.
36 ft×(1 yd/3 ft)=12 yd
Our final answer is A=1, B=3, C=12.
Answer:
A=1,000, B=1, C=5,400
Explanation:
the question was 5.4L x AmL / BmL = CmL
Given the following values of pKa, determine which is the weakest base of the answers listed. Acid pKa HClO2 1.95 HClO 7.54 HCOOH 3.74 HF 3.17 HNO2 3.15
Answer:
HClO 7.54
Explanation:
Hypochlorous acid (HClO) is a weakest acid because the pKa value of Hypochlorous acid is very high among the options given in the activity. pKa is a method which is used in order to identify the strength of an acid. The higher the value of pKa of a liquid, lower the strength of an acid while lower the value of pKa of chemical, higher the strength of an acid. In the options, HClO2 is a strong acid due to high lower pKa value.
If we represent the equilibrium as:...N2O4(g) 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature is increased the equilibrium concentration of NO2: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.
Answer:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases.
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Explanation:
Hello,
In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.
With the aforementioned, we can conclude that the chemical reaction:
[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]
Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:
[tex]G=H-TS[/tex]
As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:
[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Regards.
H2S(g) 2H2O(l)3H2(g) SO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.60 moles of H2S(g) react at standard conditions.
Answer: [tex]\Delta S[/tex] = 473.92J/K.mol
Explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:
[tex]\Delta S = m\Sigma product - n\Sigma reagent[/tex]
The balanced reaction:
[tex]H_{2}S_{(g)}+2H_{2}O_{(l)}=>3H_{2}_{(g)}+SO_{2}_{(g)}[/tex]
gives the proportion reagents react to form products, so, if 1.6 moles of [tex]H_{2}S_{(g)}[/tex]:
3.2 moles of water is used;
4.8 moles of hydrogen gas is formed;
1.6 moles of sulfur dioxide is also formed;
Calculating entropy change:
[tex]\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)[/tex]
[tex]\Delta S=628.8+398.08-328.96-224[/tex]
[tex]\Delta S[/tex] = 473.92J/K.mol
Entropy change for the given chemical reaction is [tex]\Delta S[/tex] = 473.92J/K.mol
Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.
Answer:
pH = 3.49
Explanation:
We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH of a buffer ssytem using the Henderson-Hasselbach equation.
pH = pKa + log [base] / [acid]
pH = -log Ka + log [NO₂⁻] / [HNO₂]
pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M
pH = 3.49
The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49
We'll begin by calculating the the pKa of acid. This can be obtained as follow:
Acid dissociation constant (Ka) = 4.50×10¯⁴
pKa =?pKa = –Log Ka
pKa = –Log 4.50×10¯⁴
pKa = 3.35Finally, we shall determine the pH of the solution.pKa = 3.35
Concentration of HNO₂, [HNO₂] = 0.210 M
Concentration of KNO₂, [KNO₂] = 0.290 M
pH =?The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:
pH = pKa + log [base] / [acid]pH = pKa+ log [NO₂⁻] / [HNO₂]
pH = 3.35 + log (0.290 / 0.210)
pH = 3.49Thus, the pH of the solution is 3.49
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Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?
Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
Rate constant:
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
K = 0.02295 years⁻¹
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value1. What volume in milliliters of 0.100 M HClO₃ is required to neutralize 40.0 mL of 0.140 M KOH? 2. A 25.0 mL solution of HNO₃ is neutralized with 15.7 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?
Answer:
The correct answer is 1) 56 ml and 2) 0.314 M
Explanation:
1. The reaction taking place in the given case is,
HClO₃ + KOH ⇒ KClO₃ + H2O, the molarity of HClO₃ given is 0.100 M, the molarity of KOH given is 0.140 M and the volume of KOH given is 40 ml, there is a need to find the volume of HClO₃.
Therefore, the mole of HClO₃ = mole of KOH
= MHClO₃ × VHClO₃ = MKOH × VKOH
= 0.100 M × VHClO₃ = 0.140 M × 40 ml
VHClO₃ = 0.140 M × 40 ml/0.100 M
VHClO₃ = 56 ml.
2. The reaction taking place is,
2HNO₃ + Ba(OH)₂ ⇒ Ba(NO₃)₂ + 2H₂O
The volume of HNO₃ given is 25 ml, the molarity of Ba(OH)2 is 0.250 M, the volume of Ba(OH)2 is 15.7 ml, the n or the number of moles of HNO₃ is 2, and the n of Ba(OH)2 is 1, the concentration or M of HNO₃ is,
M₁V₁/n₁ = M₂V₂/n₂
M₁ × 25/ 2 = 0.25 × 15.7/1
M₁ or molarity of HNO₃ = 0.314 M
1. The volume of HClO₃ required to neutralize the KOH is 56.0 mL
2. The concentration of the original HNO₃ solution is 0.314 M
1.
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
HClO₃ + KOH → KClO₃ + H₂O
This means,
1 mole of HClO₃ is required to neutralize 1 mole of KOH
From the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
Where
[tex]C_{A}[/tex] is the concentration of acid
[tex]C_{B}[/tex] is the concentration of base
[tex]V_{A}[/tex] is the volume of acid
[tex]V_{B}[/tex] is the volume of base
[tex]n_{A}[/tex] is the mole ratio of acid
[tex]n_{B}[/tex] is the mole ratio of base
From the given information,
[tex]C_{A} = 0.100 \ M[/tex]
[tex]C_{B} = 0.140 \ M[/tex]
[tex]V_{B} = 40.0 \ mL[/tex]
From the balanced chemical equation
[tex]n_{A} = 1[/tex]
[tex]n_{B} =1[/tex]
Putting the values into the formula, we get
[tex]\frac{0.100 \times V_{A} }{0.140 \times 40.0} = \frac{1}{1}[/tex]
∴ [tex]0.100 \times V_{A} = 0.140 \times 40.0[/tex]
[tex]V_{A}=\frac{0.140\times 40.0}{0.100}[/tex]
[tex]V_{A}=\frac{5.60}{0.100}[/tex]
[tex]V_{A}=56.0 \ mL[/tex]
Hence, the volume of HClO₃ required to neutralize the KOH is 56.0 mL
2.
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O
This means, 2 mole of HNO₃ is required to neutralize 1 mole Ba(OH)₂
From the given information,
[tex]V_{A} = 25.0\ mL[/tex]
[tex]C_{B} = 0.250 \ M[/tex]
[tex]V_{B} = 15.7 \ mL[/tex]
From the balanced chemical equation
[tex]n_{A} = 2[/tex]
[tex]n_{B} =1[/tex]
Also, Using the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
We get
[tex]\frac{C_{A} \times 25.0 }{0.250 \times 15.7} = \frac{2}{1}[/tex]
Then,
[tex]C_{A} = \frac{2\times 0.250 \times 15.7} {1 \times 25.0}[/tex]
[tex]C_{A} =\frac{7.85}{25.0}[/tex]
[tex]C_{A} =0.314 \ M[/tex]
Hence, the concentration of the original HNO₃ solution is 0.314 M
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Which molecule is NOT hypervalent?
Select the correct answer below:
SF
PBr3
PBr5
XeFo
Answer:
PBr3 is NOT hypervalent
Explanation:
The molecule that is not hypervalent is PBr3
A molecule can be defined as the smallest part of a substance that can exist independently.
It is formed by the chemical combination of two or more atoms.
A molecule is said to be hypervalent when more than four pairs of electrons are around the central atom.
A molecule is said to be hypovalent when less than four pairs of electrons are around the central atom.
From the question, the molecule that is hypovalent is PBr3
This is because, phosphorus can make hypervalent compounds, but in this specific example it is sharing three bonds and has one lone pair, so it has simply a full octet.
Therefore, the molecule that is not hypervalent is PBr3.
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How many unit cells share an atom that is located at the center of a cube edge of a unit cell?
Answer:
zero
Explanation:
In a unit cell, an atom that is located at the center of a cube edge is not involved in sharing unit cells because a central atom of a unit cell belongs to the entire cell and only to that unit cell of the lattice.
Hence, the center atom of a unit cell do not share any unit cell and the correct answer is "Zero".
15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon is 12.01
and the atomic weight of chlorine is 35.45.
O A. 11.4 moles
O B.3.53 moles
C. 5.42 moles
D. 8.35x10 moles
Answer:
well, first off. the formula for carbon tetrachloride is CCl4
We need to find the molar mass of carbon tetrachloride
1(Mass of C) + 4(mass of chlorine)
1(12) + 4(35.5)
12 + 142
154 g/mol
Number of moles of CCl3 in 543.2g CCl3
n = given mass / molar mass
n = 543.2/153
n = 3.53 moles
always remember to brainly the questions you find helpful
Answer:
3.53 moles
Explanation:
How many moles of barium sulfate are produced from 0.100 mole of barium chloride?
Answer:
0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.
Explanation:
Barium chloride and sodium sulfate react according to the following balanced reaction:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and products participate in the reaction:
BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl : 2 molesThen you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?
[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]
amount of moles of BaSO₄= 0.100
0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.
A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was ________ mm Hg.
Answer:
96 mmHg
[tex]h=96mmHg[/tex]
Explanation:
From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .
The difference of height of the mercury level is given as,
h=9.60cm
h=9.60(10mm/1cm)
[tex]h=96mm[/tex]
But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height
P(gas)=h
P(gas)=96mmHg
This pressure is as a result of the presence of gas.
Therefore, the pressure of the argon gas in the container is 96mmHg.
The pressure of the argon in the container was 96mmHg.
We were told that the manometer has closed ends which means that the
height will not be affected by atmospheric pressure.
The height which is the difference in mercury level is
h=9.60cm
We can convert it to millimeter by multiplying it by 10
h=9.60 × 10 = 96mm
The pressure of the closed end manometer will be equal to the height
P(gas)=h
P(gas)=96mmHg
The pressure of the argon gas in the container is 96mmHg.
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4. Given that the enthalpy of reaction for a system at 298 K is -292 kJ/mol and the entropy for that system is 224 J/mol K, what's the free energy for the system?
A.-87,793 kJ
B.-358 kJ
C.-225 kJ
D. -66,751 kJ
Answer:
[tex]\Delta G=-359\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:
[tex]\Delta G=\Delta H -T\Delta S\\[/tex]
In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:
[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]
Best regards.
Answer:
B- 358 kj
Explanation: I took the test
How much work (in Joules) is required to expand the volume of a pump from 0.00 L to 2.50 L against an external pressure of 1.10 atm
Answer:
W = 278.64375 Joules
Explanation:
The information given in this problem are;
Initial volume = 0L
Final volume = 2.50L
ΔV = 2.50 - 0 = 2.50 L
External pressure, P = 1.10 atm
Work = ?
These parameters are related by the equation;
w = - P ΔV
W = - (1.10 )(2.50)
W = 2.75 L atm
Upon conversion to joules;
1 liter atmosphere is equal to 101.325 joule
W = 278.64375 Joules
What is the molecular mass of this substance? a. 31.02 u b. 63.02 u c. 47.02 u d. 126.04 u e. 110.01 u
Answer:
Molecular mass of HNO₃ = 63.015 g/mol
Explanation:
Note: The given question is incomplete , the given substance is nitric acid
Given:
Nitric acid (HNO₃)
Find:
Molecular mass
Computation:
Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)
Molecular mass of HNO₃ = 1.008 + 14.007 + 48
Molecular mass of HNO₃ = 63.015 g/mol
How many grams is 5.8 moles of hydrochloric acid (HCI)?
Answer to the nearest 0.01 g.
Answer:
211.47 grams
Explanation:
We need to set up a dimensional analysis to solve this problem by converting from moles to grams.
First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:
1.01 + 35.45 = 36.46 g/mol
We have 5.8 moles of HCl, so multiply by its molar mass:
(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g
The answer is thus 211.47 grams.
~ an aesthetics over
Answer:
[tex]\large\boxed{211.47}\\[/tex] grams
Explanation:
First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.
Hydrogen has an atomic weight of 1.00784 and Chlorine has an atomic mass of 35.453.Add those two values together - 1.00784 + 35.453 = 36.46084Multiply this value by 5.8 (one mole is equivalent to the atomic mass of the compound) - 5.8 x 36.46084 = 211.472872Round to the nearest 0.01 gram - 211.47[tex]\large\boxed{211.47}[/tex] is the final answer.
What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)
Answer: d=2000 g/L
Explanation:
Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.
[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]
Now that we have grams, we can divide to get density.
[tex]d=\frac{10000g}{5 L}[/tex]
d=2000g/L
Arrange the following substances in the order of increasing entropy at 25°C. HF(g), NaF(s), SiF 4(g), SiH 4(g), Al(s) lowest → highest
Answer:
Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)
Explanation:
Hello,
In this case, we can arrange the increasing order of entropy at 25 \°C by taking into account, at first, that since solids are more molecularly organized than gases, the first we have solid sodium fluoride and solid aluminium, but in this case, as the higher the molar mass, the higher the entropy, the molar mass of aluminium is 27 g/mol and 42 g/mol for sodium fluoride, therefore, we first have:
Al(s)<NaF(s)
Afterwards, since the molar mass of hydrogen fluoride (HF), silicon fluoride (SiF4) and silane (SiH4) are 20, 104 and 32 g/mol respctively, since silicon fluoride has the greater molar mass, it also has the higher entropy. In such a way, the overall order turns out:
Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)
Best regards.
PLEASE HELP!! 40 POINTS
Answer:
1) 6.524779402×10^(-17)
2)521.1g
3)113
Explanation:
Answer: 1) 6.524779402×10^(-17)
2)521.1g
Explanation:
Human blood typically contains 1.04 kg/L of platelets. A 1.89 pints of blood would contain what mass (in grams) of platelets
A 1.89 pints of blood would contain 873 grams of platelets.
To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:
1 pint = 473.2 mL[tex]1.89 \times 473.2 = 894.3 mL[/tex]
1000 L = 1mL
[tex]\frac{894.3}{1000}= 0.84L[/tex]
Now, just calculate the amount of platelets present in 0.84L:
[tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]
x = 873 grams
So, a 1.89 pints of blood would contain 873 grams of platelets.
Learn more about transformation of units in: brainly.com/question/10667910
how can you prevent frequent landslides from heavy rains
Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.
Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud
B. Plant vegetation on the slops that expreience landslides.
E. Reduce the sloped where landslides occur.
Hope this helped, please mark brainiest!
Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 428.7 pm, what is the density of CsBr in g/cm3.
Answer:
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
Explanation:
Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.
The density d of CsBr in g/cm³ can be calculated by using the formula:
[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]
where;
z = 1 mole of CsBr
edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm
molar mass of CsBr = 212.81 g/mol
avogadro's number = 6.023 × 10²³
[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]
[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]
[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]
When the nuclide bismuth-210 undergoes alpha decay:
The name of the product nuclide is_____.
The symbol for the product nuclide is_____
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_____ rightarrow 4He+ 234Th
2 90
Write a balanced nuclear equation for the following:
The nuclide radium-226 undergoes alpha emission.
Explanation:
An atom undergoes alpha decay by losing a helium atom.
So when bismuth undergoes alpha decay, we have;
²¹⁰₈₃Bi --> ⁴₂He + X
Mass number;
210 = 4 + x
x = 206
Atomic number;
83 = 2 + x
x = 81
The element is Thallium. The symbol is Ti.
For the second part;
X --> ⁴₂He + ²³⁴₉₀Th
Mass number;
x = 4 + 234 = 238
Atomic Number;
x = 2 + 90 = 92
The balanced nuclear equation is;
²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th
Which of the following processes release energy? a. ball rolling down a hill b. formation of copper (II) oxide from copper and oxygen c. formation of ice from liquid water d. condensation of water on a wind shield of a car
Answer:
d. condensation of water on a wind shield of a car
Explanation:
Condensation involves the conversion of moist air into liquid.
Gas has a higher energy compared to liquid. This is why Gas particles move at random motion and faster in relation to solid and liquid particles due to the high energy content.
The conversion of the gas to liquid means that there was loss or release of energy which validates the answer.
Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b 2 ( a q ) | P b ( s ) Al(s)|AlX3 (aq)||PbX2 (aq)|Pb(s)
The question is missing. Here is the complete question.
Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex]?
(a) [tex]Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}[/tex]
(b) [tex]2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}[/tex]
(c)[tex]Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}[/tex]
(d) [tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
Answer: (d) [tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.
Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.
For the cell notation: [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex]
Aluminum's half-equation is oxidation:
[tex]Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}[/tex]
For Lead, half-equation is reduction:
[tex]Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}[/tex]
Multiply first half-equation for 2 and second half-equation by 3:
[tex]2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}[/tex]
[tex]3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}[/tex]
Adding them:
[tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
The balanced redox reaction with cell notation [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex] is
[tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
