Answer:
Explanation:
The elements in Group 1A elements are Lithium, Sodium, Potassium, Rubidium, Caesium, and Francium(radioactive). The valence shells of the members have the electronic configuration of ns¹ where n =2,3,4,5,6,7. They are known as alkali earth metals. Their chemistry is dominated by the only electron in the outer shell of their atoms. Thus, they are powerful reducing agents because of their ability to lose this outer shell electron readily.
Thus ns' electrons in Group 1A require lesser energy for ionization in the case of elements of Group 1A; this explains why Group 1A elements are more reactive.
On the other hand, the Group 1B elements outer ns' electrons are less effectively shielded by the inner d electrons; therefore, the outer electrons in case of Group 1B feel more nuclear pull (due to positively charged protons in the nucleus) as compared to the nuclear pull in Group 1A elements. Thus electrons require more energy for ionization in the case of elements of Group 1B. This explains why Group 1B elements are less reactive than elements of Group 1A.
what is the concentration of Hg2 in equilibrium with 0.10M Cl- in a solution of KCl containing excess undissolved Hg2Cl2(s)
Answer:
[Hg₂²⁺] = 1.3 × 10⁻¹⁶ M
Explanation:
Step 1: Given data
[Cl⁻]: 0.10 M
Solubility product constant of Hg₂Cl₂ at 25 °C (Ksp): 1.3 × 10⁻¹⁸
Step 2: Write the reaction for the solution of mercury (I) chloride
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
Step 3: Use the solubility product constant to calculate [Hg₂²⁺]
Ksp = 1.3 × 10⁻¹⁸ = [Hg₂²⁺].[Cl⁻]² = [Hg₂²⁺].(0.10)²
[Hg₂²⁺] = 1.3 × 10⁻¹⁶ M
How many grams of water can be heated from 25.0 C to 35.0 C by the heat released from 85.0 g of iron that cools from 85 C to 35 C
Answer:
31.94 g
Explanation:
Water
ΔT = 35 C - 25 C = 10
m = ?
c = 4.182 J/gC
Heat = mc ΔT
Iron
ΔT = 85 C - 50 C = 35 (Negative sign is not relevant)
m = 85
c = 0.449 J/gC
Heat = mc ΔT= 85 * 0.449 * 35 = 1335.76 J
Substituting the value of heat in the equation for water;
Heat = m * 4.182 * 10
1335.76 = m * 41.82
m = 31.94 g
The image on the left shows the charges on a balloon after it’s been rubbed with a wool cloth. The image on the right is a piece of tissue paper. What will happen to the tissue paper as it approaches the balloon and why?
Answer:
B = tissue paper will stick to the balloon because it is positively charged.
Explanation:
The option B is correct. Balloon is negatively charged while it is stated in question and can also see from the picture tissue paper is positively charged.
We know that opposite charges attract each other. Thus when balloon and tissue paper come close to each other the stick together.
All other options are incorrect because,
Option A:
In option it is stated that tissue paper is negatively charged. Although negative charges are present but there is more positive charge in balloon we can say that more number of protons are present thus balloon have positive charge.
Option C:
Tissue paper is negatively charged and will move away.
It is wrong. Tissue paper is positively charged.
Option D:
Tissue paper will move away because it is positively charged. It is also wrong statement because tissue paper will not move away it will stick to the balloon because of positive charge.
A given sample of a gas has a volume of 3.0 L at a pressure of 4.0 atm. If the temperature remains constant and the pressure is Changed to 6.0 atm, why is the new volume?
Answer:
2L
Explanation:
P1xV1=P2xV2
P1=4 atm
V1=3L
P2=6 atm
V2= ?
V2= (4 x 3)/ 6
V2=2L
What are the major characteristics of metals
Describe how your response in the previous question explain the decrease in the temperature measured during the evaporation process
Answer:
See explanation
Explanation:
Evaporation is an endothermic process. In an endothermic process, heat is absorbed from the surrounding. This causes the system to feel cool.
Since evaporation is endothermic and heat is absorbed from the surrounding, the system will feel cool and the temperature of the system decreases accordingly.
Hence, during evaporation, the temperature of the system decreases.
Determine the approximate density of a high-leaded brass that has a composition of 60.5 wt% Cu, 34.5 wt% Zn, and 5.0 wt% Pb. The densities of Cu, Zn, and Pb are 8.94, 7.13 and 11.35 g/cm3, respectively.
Answer:Approximate density of the high-leaded brass(alloy) =8.306g/cm³
Explanation:
The density of an alloy is its mass (100g) divided by its volume
Therefore we have that the alloy (high-leaded brass) has a composition of
60.5 wt% Cu with density 8.94g/cm3
34.5 wt% Zn, wth densty 7.13g/cm3
5.0 wt% Pb with densty g/cm3
The total volume of the alloy will be the mass / density of ts composition given as :
60.5gCu/DCu + 34.5gZn/DZn + 5.0gPb/Dpb
= 60.5/8.94 + 34.5/7.13+ 5.0/11.35
= 6.76 cm³ + 4.838 cm³+0.4405 cm³ = 12.0385cm³
Approximate density of the high-leaded brass(alloy) = 100g/ 12.0385cm³ =8.306g/cm³
The diagram shows the different models of the atom that eventually led to the modern atomic theory.
Which model represents the current atomic theory based on our knowledge of the nucleus and electron cloud?
1) Model A
2) Model B
3) Model C
4) Model D
The model depicted that has been resembling the modern atomic theory has been Model D. Thus option 4 is correct.
The atom has been the smallest unit of the matter, that has been consisted of the electrons, protons, and neutrons as the subatomic particles.
The modern atomic theory has stated that the structure of the atom has a positively charged centrally located nucleus, with the negatively charged body revolving around it.
The central nucleus has consisted of the neutral neutrons and positive protons, while the revolving bodies have been the negative electrons. The model depicted that has been resembling the modern atomic theory has been Model D. Thus option 4 is correct.
For more information about the modern atomic theory, refer to the link:
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Answer: Model D; answers 4
Explanation: Is the most modern and recent atom model based on our knowledge.
Write a formula for the compound that forms from potassium and acetate. Express your answer as a chemical formula. Write a formula for the compound that forms from potassium and chromate.
Answer:
KCH₃COO
K₂CrO₄
Explanation:
Write a formula for the compound that forms from potassium and acetate.
Potassium has a charge +1 (K⁺) and acetate has a charge -1 (CH₃COO⁻). Thus, the neutral compound is KCH₃COO (potassium acetate).
Write a formula for the compound that forms from potassium and chromate.
Potassium has a charge +1 (K⁺) and chromate has a charge -2 (CrO₄²⁻). Thus, the neutral compound is K₂CrO₄ (potassium chromate).
Reactivity of a_______ depends on how easily an electron(e-) can be removed. (This is lonization energy)
Answer:
1. How is ionization energy related to how easily an electron can be lost from an atom? The easier it is to lose an electron the smaller the amount of energy needed to remove it (smaller ionization energy). ... The amount of positive charge increases as the number of protons increase from left to right across a period.
Explanation:
You turn on the radio in the middle of a report about a planet. You don’t hear the planet’s name, but you hear that it is a large, gas giant with 66 confirmed moons and a Great Red Spot. Based on this information, which planet is being discussed?
Group of answer choices
Neptune
Uranus
Saturn
Jupiter
Answer:
Jupiter
Explanation:
What molecules from food and air end up in the cells in the body?
PLEASE
Answer:
Molecules from food and molecules of oxygen move from the mouth and the nose to cells of the body through a series of blood vessels, including veins, arteries, and microscopically small blood vessels (capillaries), that extend throughout the body.Explanation:
Answer:
Molecules from food and air enter your body through your nose and mouth. These particles go down your throat and your lungs, and to your stomach. The blood also distributes these molecules.
Hope this helps!
To make a new solution, 35.0 mL of a 12.0 M HCl stock solution is diluted to a volume of 1.20 L. What is the concentration of the new solution?
Answer:
m1v1=m2v2 change 35 ml to l = 0.035L
12*0.035 = M2*1.20
Solve for M2 = 0.35M
Explanation:
What kind of bond would two nonmetals typically make?
A. An ionic bond
B. A covalent bond
C. A metallic bond
D. An ionic solid
Answer:
covalent bond
Explanation:
At high temperatures, alkanes can undergo dehydrogenation to produce alkenes. For example: This reaction is used industrially to prepare ethylene while simultaneously serving as a source of hydrogen gas. Explain why dehydrogenation only works at high temperatures.
Answer:
Dehydrogenation of alkanes is endothermic
Explanation:
The dehydrogenation of alkanes is an endothermic reaction. The enthalpy change for the reaction is highly positive.
Recall that for an endothermic reaction, increase in temperature increases the rate of forward reaction.
As a result of that, the dehydrogenation of alkanes proceeds in the foward direction at elevated temperatures.
How many moles are in 36 g of Be?
(NH4)3PO4 + CaBr2 = Ca3(PO4)2 + NH4Br
ionic equation
Answer:
2(NH4)3PO4 + 3CaBr2 → Ca3(PO4)2 + 6NH4Br
Explanation:
Tribasic Ammonium Phosphate + Calcium Bromide = Tricalcium Phosphate + Ammonium Bromide
Reaction Type
Double Displacement (Metathesis)
Reactants
Tribasic Ammonium Phosphate - (NH4)3PO4
Ammonium Phosphate
Molar Mass of H12N3O4P Oxidation State of H12N3O4P
Calcium Bromide - CaBr2
Kalziumbromid CaBr2 Calcium Bromide Hydrate
Molar Mass of Br2Ca Oxidation State of Br2Ca
Products
Tricalcium Phosphate - Ca3(PO4)2
Bone Phosphate Of Lime BCP TCP Tricalcium Orthophosphate Calcium Phosphate Tertiary Calcium Phosphate Calcium Orthophosphate Ca3(PO4)2
Molar Mass of Ca3O8P2 Oxidation State of Ca3O8P2
Ammonium Bromide - NH4Br
NH4Br
Molar Mass of BrH4N Oxidation State of BrH4N
If a 0.775 M solution of a base is found to have a hydroxide concentration of 0.050 M at equilibrium, what is the percent ionization of the base
Answer:
[tex]\%\ ionization=6.45\%[/tex]
Explanation:
Hello.
In this case, since the ionization of a hypothetical base MOH is:
[tex]MOH(aq)\rightleftharpoons M^+(aq)+OH^-(aq)[/tex]
The equilibrium expression is:
[tex]Kb=\frac{[M^+][OH^-]}{[MOH]}[/tex]
And the percent ionization is:
[tex]\%\ ionization=\frac{[OH^-]}{[MOH]} *100\%[/tex]
For the equilibrium concentration of hydroxide, it turns out:
[tex]\%\ ionization=\frac{0.050M}{0.775M}*100\%\\\\\%\ ionization=6.45\%[/tex]
Best regards!
3. How are the three mutated strands different from each other?
Answer:
Two major categories of mutations are germline mutations and somatic mutations. Germline mutations occur in gametes. These mutations are especially significant because they can be transmitted to offspring and every cell in the offspring will have the mutation. Somatic mutations occur in other cells of the body
Explanation:
g o o g l e
Two major categories of mutations are germline mutations and somatic mutations. Germline mutations occur in gametes.
What is mutation?A mutation is a change to an organism's DNA sequence. Errors in DNA replication during cell division, exposure to mutagens, or viral infection can all cause mutations. Somatic mutations (which happen in body cells) cannot be passed on to offspring, whereas germline mutations (which happen in eggs and sperm) can.
Our cells are constantly undergoing mutations, yet very few of these have any negative effects on our health. Compared to what we typically see in science fiction movies, this is completely different.
A mutation is never so advantageous in real life that it makes someone a superhero or does anything strange like give them wings.
Therefore, Two major categories of mutations are germline mutations and somatic mutations. Germline mutations occur in gametes.
To learn more about gametes, refer to the link:
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Compute the bond energy of the C-Cl bond using the reaction and data in question 10, the value of the bond energy for the C-H bond of 410 kJ/mol and the value of the H-H bond of 432 kJ/mol. The C-Cl bond energy in kJ/mol is:
Answer:
The bond formation of C-Cl bond by applying the given data in question 10 is 325.4 kJ
Explanation:
From the data in question 10:
The equation for the given reaction can be expressed as:
[tex]CH_{4(g)} + 2Cl_{(g)} \to CH_2Cl_2_{(g)}+H_2_{(g)}[/tex]
The enthalpy of formation of the compounds are as follows:
Substance [tex]\Delta \ H^0_f \ (kJ/mol)[/tex]
[tex]CH_{4(g)}[/tex] - 74.8
[tex]Cl_{(g)}[/tex] 120.9
[tex]CH_2Cl_{2(g)}[/tex] -95.8
From the data obtained above, the enthalpy of the reaction [tex]\Delta \ H ^0_{rxn}[/tex] can be computed as follows:
[tex]\Delta H^0_{rxn } = \sum ( \Delta H^0_f )_{products} - \sum ( \Delta H^0_f )_{reactants}[/tex]
[tex]\Delta H^0_{rxn } = [ 1 \ mol \times ( \Delta H^0_{CH_2Cl_{2(g)}}) + 1 \ mol \times ( \Delta H^0_{H_{2(g)}})] - [ 1 mol \times (\Delta H^0_{CH_4}_{(g)} +2 \ mol \times (\Delta \ H^0 _{Cl(g)})]}[/tex]
[tex]\Delta^0_{rxn } = [ 1 \ mol \times ( -95.8 \ kJ/mol) + 1 \ mol \times ( 0 \ kJ/mol] - [ 1 mol \times (-74.8 \ kJ/mol +2 \ mol \times (120.9 \ kJ/mol)]}[/tex]
[tex]\Delta H^0_{rxn } =-262.8 \ kJ[/tex]
However; since we knew [tex]\Delta H^0_{rxn }[/tex] to be -262.8 kJ
At standard conditions
Bond Bond Energy (kJ/mol)
C-H 410
H-H 432
The bond energy can be calculated by using the expression:
[tex]\Delta H^0_{rxn } = \sum \Delta H _{bond \ broken }- \sum \Delta H _{bond \ formed}[/tex]
[tex]-262. 8 KJ =[4 \times ( \Delta H_{CH})] - 1 [ (2 \times \Delta H_{CH} ) + ( 2 \times \Delta H_{C-Cl}) + (1 + \Delta H _{H-H})][/tex]
[tex]-262. 8 KJ =[4 \times ( 410 \ kJ/mol) ] - 1 [ (2 \times 410 \ kJ/mol ) + ( 2 \times \Delta H_{C-Cl}) + (1 +432 \ kJ/mol][/tex]
[tex]-262. 8 KJ = 388 \ kJ - ( 2 \times \Delta H_{C-Cl})[/tex]
[tex]( 2 \times \Delta H_{C-Cl}) = 388 \ kJ +262. 8 KJ[/tex]
[tex]( 2 \times \Delta H_{C-Cl}) = 650.8 \ KJ[/tex]
[tex]\Delta H_{C-Cl} = \dfrac{650.8 \ KJ}{2}[/tex]
[tex]\Delta H_{C-Cl} = 325.4 \ kJ[/tex]
Thus; the bond formation of C-Cl bond by applying the given data in question 10 is 325.4 kJ
the equilibrium constant, K, is greater than one. The stronger acid in this system is ___________ and the weaker base is ____________.
The question is incomplete; the complete question is;
For the following reaction:
H2CO3 + S2- ---------> HCO3- + HS-
The equilibrium constant is greater than one. The strongest acid in this system is ___________ and the weakest base is ____________.
A) H2CO3 and HS-
B) S2- and HS-
C) HCO3- and S2-
D) HS- and S2-
E) H2CO3 and HCO3-
Answer:
H2CO3 and HS-
Explanation:
According to the Brownstead- Lowry theory of acids and bases, an acid is a proton donor while a base is specie that accepts protons.
A strong acid has a greater tendency to donate protons. If we look at the reaction, H2CO3 has the greatest ability to donate protons hence it is the strongest acid.
On the other hand, HS- has the least ability to accept protons so it is the weakest base.
Butane (C4H10) burns in the presence of oxygen to produce carbon dioxide gas and water vapor. Write a balanced equation for this reaction and calculate the number of liters of carbon dioxide measured at STP that could be produced from 5.69g of butane
Answer:
8.78 L
Explanation:
The balanced equation for the combustion of butane (C₄H₁₀) in presence of oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) is the following:
2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O
First, we calculate the moles there are in 5.69 g of butane, by dividing the mass into the molecular weight (MM butane= 58 g/mol):
moles C₄H₁₀= mass/MM= 5.69 g/(58 g/mol)= 0.098 mol
According to the equation, 2 moles of C₄H₁₀ produce 8 moles of CO₂. Thus, if we have 0.098 moles C₄H₁₀:
moles CO₂= 0.098 mol C₄H₁₀ x 8 mol CO₂/(2 mol C₄H₁₀)= 0.392 mol CO₂
Finally, we know that 1 mole of any gas occupies a volume of 22.4 L measured at STP, so we calculate the volume of CO₂ as follows:
volume CO₂= 0.392 mol CO₂ x 22.4 L/1 mol CO₂= 8.78 L
convert 3.01•10^23 molecules of C2H6 to moles.
Answer:
The answer is 0.5 molesExplanation:
To find the number of moles in a substance given it's number of entities we use the formula
[tex]n = \frac{N}{L} \\ [/tex]
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
[tex]n = \frac{3.01 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{3.01}{6.02} \\ [/tex]
We have the final answer as
0.5 molesHope this helps you
When 0.627 g of biphenyl (CH) combusts in a bomb calorimeter, the temperature rises from 21.9 C to 26.6 C. What is U for the reaction in kJ/mol biphenyl
Answer:
[tex]U=-6852\frac{kJ}{mol}[/tex]
Explanation:
Hello there!
In this case, since the heat capacity of the bomb calorimeter, we can use this found value on ethernet: 5.86 kJ/°C, thus, we can compute the heat gained by the calorimeter:
[tex]Q_{calorimeter}=C\Delta T=5.86kJ/\°C*(26.6-21.9)\°C=27.54kJ[/tex]
Thus, since the heat gained by the calorimeter is actually the heat released due to the combustion of byphenyl, we compute it as the negative of that of the calorimeter:
[tex]Q_{combustion}=-Q_{calorimeter}=-27.54kJ[/tex]
Finally the energy of reaction, U, in kJ/mol for the combustion of byphenyl (molar mass = 154 g/mol) is:
[tex]U=\frac{-27.54kJ}{0.627g*\frac{1mol}{156g} } \\\\U=-6852\frac{kJ}{mol}[/tex]
Best regards!
g Determine the percent enantiomeric excess of a sample with specific rotation of 14 compared to an enantiomerically pure reference stereoisomer of the sample that has specific rotation -24
Answer:
Explanation:
Enantiomeric excess of this compound X is given by the formula
EE= Observed specific rotation*100 / maximum specific rotation
EE= 14*100/24 = 58.33%
Percentage enantiomeric excess
%(-) = EE/2 +50
= 58.33/2 + 50
= 79.12% OF (-)X and 20.88% of (+) X
How many moles of H2 can be formed if a 2.97 g sample of Mg reacts with excess HCl?
The number of moles of H₂ formed when 2.97 g sample of Mg reacts with excess HCl is 0.122 moles
To determine the number of moles of H₂ that can be formed if 2.97 g sample of Mg reacts with excess HCl,
First, we will write a balanced chemical equation for the reaction
Mg + 2HCl → MgCl₂ + H₂
From the reaction above,
It means that 1 mole of Mg will react with 2 moles of HCl to produce 1 mole of MgCl₂ and 1 mole of H₂
Now, we will determine the number of moles of Mg present in the given sample.
From the question,
Mass of Mg = 2.97 g
Using the formula
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]
Molar mass of Mg = 24.305 g/mol
∴ [tex]Number \ moles \ of \ Mg = \frac{2.97}{24.305}[/tex]
Number of moles of Mg in the sample = 0.122 moles
Since, 1 mole of Mg will react with 2 moles of HCl to produce 1 mole of MgCl₂ and 1 mole of H₂
∴ 0.122 moles of Mg will react with excess HCl to produce 0.122 moles of H₂
Hence, the number of moles of H₂ formed when 2.97 g sample of Mg reacts with excess HCl is 0.122 moles
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Squalene, an important intermediate in the biosynthesis of steroids, has the molecular formula C30H50 and has no triple bonds. (a) What is the index of hydrogen deficiency of squalene
Answer:
6
Explanation:
Molecular Formular = C30H50
Degree of saturation = (2C + 2 - H ) / 2
Degree of saturation = [ 2(30) + 2 - 50 ] / 2
Degree of saturation = ( 62 -50 ) / 2 = 12 / 2 = 6
Enter your answer in the provided box. Find the pH of a buffer that consists of 0.31 M boric acid (H3BO3) and 0.36 M sodium borate (NaH2BO3). (pKa of boric acid
Answer:
pH = 9.30
Explanation:
We can obtain the pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pH is the pH of the solution,
pKa the pKa of the buffer (9.24 for boric acid)
[A⁻] the molarity of the conjugate base, NaH2BO3
[HA] molarity of the weak acid, H3BO3
Replacing:
pH = 9.24 + log [0.36M] / [0.31M]
pH = 9.30
Why do you think it’s important to follow the steps of the scientific method?
i think science is important bcoz it has a choice whether to save or not
like now scientist are finding a cure for corona
The reference compound for naming D and L isomers of sugars is: A. glucose. B. ribose. C. glyceraldehyde. D. fructose. E. sucrose.
Answer:
C
Explanation:
The reference compound for naming D and L isomers of sugar is glyceraldehyde.
D and L stand for dexter (right) and laevus (left). The D and L system references molecules to that of glyceraldehyde which is chiral. By chirality, it means that glyceraldehyde molecules have the capacity to produce non-superposable mirror images. The images are labeled D and L isomers for 'right rotary' and 'left rotary' respectively.
This property allows some chemical manipulations to be performed on the molecules of glyceraldehyde without changing its configuration and as such, other compounds with similar ability are named by analogy to glyceraldehyde.
The correct option is C.
