Answer:
Polythene bags are so lightweight and aerodynamic, they are easily picked up and carried by the wind. They can escape from trash bins, recycle bins, garbage trucks, and landfills, and end up littering the landscape.
hope it helps
Explanation:
calculate the energy in joules and calories required to heat 50.0g silver from 106c to 255c.
Answer:
Explanation:
use this fromula
q = m c ∆t
m is mass of silver =50 g
∆t is difference in temperature= 255-106=149
C= specific heat fo silver ( should be mentioned in your question )
An electron in a 3s3s orbital penetrates into the region occupied by core electrons more than electrons in a 3p3p orbital. An electron in a orbital penetrates into the region occupied by core electrons more than electrons in a orbital. true false
Answer:
True
Explanation:
The penetrating ability of electrons in the orbitals is in the order s > p > d > f
An electron in a 3s orbital is closer to the nucleus than the one in a 3p orbital and as a result, there will be lesser shielding effect on it. This low shielding effect experienced by the 3s electron gives it a high penetration ability and hence will be able to easily penetrate regions occupied by core electrons. Conversely, the 3p orbital is farther away from the nucleus, electrons revolving around it are highly shielded which limits their ability to penetrate regions of core electrons.
Note that the maximum electrons that the s orbital can accommodate is 2 while p orbital can accommodate a maximum of 8.
You are titrating 25.025.0 mL of 0.0100 M Sn2 Sn2 in 1 M HCl HCl with 0.0500 M Tl3 Tl3 resulting in the formation of Sn4 Sn4 and Tl Tl . A PtPt indicator electrode and a saturated Ag|AgCl Ag|AgCl reference electrode are used to monitor the titration. What is the balanced titration reaction
Answer:
Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺
Explanation:
The Sn²⁺ is oxidized to Sn⁴⁺. Whereas Tl³⁺ is reduced to Tl⁺. The half-reactions are:
Sn²⁺ → 2e⁻ + Sn⁴⁺ (Oxidation, loosing electrons)
Tl³⁺ + 2e⁻ →Tl⁺ (Reduction, gaining electrons)
The sum of the reactions gives:
Sn²⁺ + Tl³⁺ + 2e⁻ → 2e⁻ + Sn⁴⁺ + Tl⁺
Subtracting the electrons in both sides of the reaction:
Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺Which diagram represents the bonding pattern of metals?
Answer:
there's no image can't help without it sorry
Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.
Question:
When 0.500 g of biphenyl (C₁₂H₁₀) undergoes combustion in a bomb calorimeter, the temperature rises from 26.8 °C to 29.5 °C. Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.
Answer;
-4870kJ/mol(3 significant figure)
475 grams of solid calcium oxide reacts with water vapor to form solid calcium hydroxide.Calculate the heat of reaction.
(A) - 221.1 kcal
(B) + 290.8 kcal
C) - 290.8 kcal
(D) + 324.6 kcal
(E) + 221.1 kcal
Answer:
(A) - 221.1 kcal
Explanation:
Based in the reaction:
CaO(s) + H₂O(g) → Ca(OH)₂ ΔH = -109kJ/mol
When 1 mole of CaO reacts per mole of water vapor producing calcium hydroxide there are released -109kJ
475g of CaO (Molar mass CaO: 56.08g/mol) are:
475g CaO × (1mol / 56.08g) = 8.47 moles of CaO
As 1 mole of CaO in reaction release -109kJ, 8.47 moles release:
8.47 mol CaO × (-109 kJ / 1 mol CaO) = -923.2kJ are released
As 1 kCal = 4.184kJ:
-923.2kJ × (1kCal / 4.184kJ) =
-220.7kCal ≈ (A) - 221.1 kcalUsing the Henderson-Hasselbalch equation, calculate the percentage of an ammonia solution that would be in the ionic form at pH 9.40, assuming a pKa for the ammonium ion of 9.25.
Answer:
41.45% is the precentage of the ionic form
Explanation:
H-H equation is used to find pH of a buffer. The formula is:
pH = pKa + log₁₀ [X⁻] / [HX]
Where X⁻ is the conjugate base of the weak acid, HX.
In the ammonia buffer:
NH₄⁺ ⇄ NH₃ + H⁺
NH₄⁺ is the weak acid (Ionic form) and NH₃ is the conjugate base. Replacing:
9.40 = 9.25 + log₁₀ [NH₃] / [NH₄⁺]
1.4125 = [NH₃] / [NH₄⁺] (1)
In a percentage, [NH₃] + [NH₄⁺] = 100 (2)
Replacing (2) in (1):
1.4125 = 100 - [NH₄⁺] / [NH₄⁺]
1.4125[NH₄⁺] = 100 - [NH₄⁺]
2.4125[NH₄⁺] = 100
[NH₄⁺] = 41.45% is the precentage of the ionic form
The percentage of an ammonia solution is 41.5%
The Henderson Hasselbalch equation is an approximation that demonstrates the relation among a solution's pH or pOH, as well as the concentration ratio for the dissociated chemical species.
The dissociation of ammonium ion in the buffer is as follow:
[tex]\mathbf{NH_4^+ \to NH_3 + H^+}[/tex]
The Henderson Hasselbalch equation for the dissociation of ammonium can be expressed by using the formula:
[tex]\mathbf{pH = pKa + log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
Given that:
pH = 9.40pKa = 9.25[tex]\mathbf{9.40= 9.25 + log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{9.40- 9.25 = log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{0.15 = log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{10^{0.15} = \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{1.41 = \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{\dfrac{141}{100} = \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
∴
The percentage of NH₄⁺ [tex]\mathbf{= \dfrac{141}{100+141}\times 100\%}[/tex]
= 58.5%
The percentage of NH₃ [tex]\mathbf{= \dfrac{100}{100+141} \times 100\%}[/tex]
= 41.5%
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Will GIVE BRAINLIEST ---A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water.
Answer the following questions, being sure to include units and remember sig figs. Show your work.
a. What is the concentration of this standard solution?
Answer:
a) 0.5198 mol/L
b) 0.00811 mol KOH
Explanation:
a) M(KOH) = 39+16+1= 56 g/mol
14.555g* 1 mol/56 g = 14.555/56 mol KOH in 500.0 mL solution
14.555/56 mol KOH ---- 0.5000 L solution
x mol KOH ----- 1 L solution
x = (14.555/56)mol * 1L/0.5000 L = 0.5198 mol/L
b) If the student pours out a 15.6 mL sample of this solution, how many moles of sodium hydroxide? (if we talk about KOH)
does the student have in the sample
15.6 mL = 0.0156 L
0.5198 mol/L * 0.0156 L = 0.00811 mol KOH
Molarity is an important method which is used to determine the concentration of a solution. The concentration of the standard solution containing 14.555 g to 500.0 mL of water is 29.11 mol L⁻¹.
What is molarity?The molarity of a solution is an important method which is used to calculate the concentration of a binary solution which contains the solute and solvent. It is defined as the number of moles of the solute present per litre of the solution.
The molarity is generally represented by the letter 'M' and its unit is mol L⁻¹. The equation used to calculate the molarity is given as:
Molarity = Number of moles of the solute / Volume of the solution in litres
500.0 mL = 0.5 L
M = 14.555 / 0.5 = 29.11 mol L⁻¹
Thus the concentration of a solution is 29.11 mol L⁻¹.
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Which is regarding enzyme inhbition
Explanation:
An enzyme inhibitor is a molecule that binds to an enzyme and decreases its activity. ... Since blocking an enzyme's activity can kill a pathogen or correct a metabolic imbalance, many drugs are enzyme inhibitors. They are also used in pesticides.
The air in a hot-air balloon at 735 torr is heated from 22.0°C to 46.0°C. Assuming that the moles of air and the pressure remain constant, what is the density of the air at each temperature? (The average molar mass of air is 29.0 g/mol.)
Answer:
D1 = 1.16 g/L
D2 = 1.07 g/L
Explanation:
PV = mRT/M
PVM = mRT
PM/RT = m/V = D - density
D = PM/RT
P = 735 torr
T1= 22.0°C+ 273.15 = 295.15 K
T2 = 46.0°C+ 273.15 = 319.15 K
M = 29.0 g/mol
R = 62.363 torr*L/mol*K
D1 = PM/RT1
D1 = (735 torr*29.0 g/mol) / (62.363 torr*L/mol*K *295.15 K) = 1.16 g/L
D1 = 1.16 g/L
D2 = PM/RT2
D2 = (735 torr*29.0 g/mol) / (62.363 torr*L/mol*K *319.15 K) = 1.07 g/L
D2 = 1.07 g/L
Which of the following aqueous solutions are good buffer systems?
0.31 M ammonium bromide + 0.39 M ammonia
0.31 M nitrous acid + 0.25 M potassium nitrite
0.21 M perchloric acid + 0.21 M potassium perchlorate
0.16 M potassium cyanide + 0.21 M hydrocyanic acid
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite
0.13 M nitrous acid + 0.12 M potassium nitrite
0.15 M potassium hydroxide + 0.22 M potassium bromide
0.23 M hydrobromic acid + 0.20 M potassium bromide
0.34 M calcium iodide + 0.29 M potassium iodide
0.33 M ammonia + 0.30 M sodium hydroxide
0.20 M nitrous acid + 0.18 M potassium nitrite
0.30 M ammonia + 0.34 M ammonium bromide
0.29 M hydrobromic acid + 0.22 M sodium bromide
0.17 M calcium hydroxide + 0.28 M calcium bromide
0.34 M potassium iodide + 0.27 M potassium bromide
Answer:
Answers are in the explanation.
Explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:
0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.
0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.
0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.
0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.
0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.
0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.
0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.
0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.
0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.
0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.
0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.
0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.
0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.
0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.
Which describes an effect that ocean currents have on short-term climate change? Ocean currents increase the strength of prevailing winds, which can cool the air and land. Ocean currents can carry cold water, which can cool the air and land. Ocean currents increase hurricane activity, which can raise the temperature of the air and land. Ocean currents can carry warm water, which causes hurricane activity and raises the temperature of the air and land.
Answer: B
Ocean currents can carry cold water, which can cool the air and land.
Explanation:
eet ees wat eet ees
plz mark brainliest
Answer:
B is right
Explanation:
which proess is part of the carbon cycle
Answer:
The key processes in the carbon cycle are: carbon dioxide from the atmosphere is converted into plant material in the biosphere by photosynthesis.
Explanation:
organisms in the biosphere obtain energy by respiration and so release carbon dioxide that was originally trapped by photosynthesis. ... The carbon becomes part of the .
How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
O A. Elements in the same group have the same number of valence
electrons.
B. The number of valence electrons increases as the atomic number
increases
C. The number of valence electrons is the same for all elements on
the periodic table.
D. Elements in the same period have the same number of valence
electrons.
Answer:
A
Explanation:
g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass. Must show your work on scratch paper to receive credit.
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
Identify each element below, and give the symbols of the other elements in its group.
a. [Ar] 4s23d104p4
b. [Xe] 6s24f145d2
c. [Ar] 4s23d5.
Answer:
Explanation:
the electron configuration is defined as the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is
used to describe the orbitals of an atom in its ground state
The valence electrons, electrons in the outermost shell, can be used to know the chemical property
a)
Chemical Name of the Element: Selenium
Chemical Symbol: Se
Group it belong in periodic table:6A
Other Element in the same group:tellurium(Te),,sulfur(S)
atomic number = 34
Selenium is a chemical element that has symbol Se It is a nonmetal which is usually classified as metalloid with properties that are intermediate between the elements above and below in the periodic table.
b)Chemical Name of the Element:Hafnium
Chemical Symbol: Hf
Group it belong in periodic table:4B
Other Element in the same group: Titanium( Ti )Rutherfordium
atomic number: 72
Hafnium is a solid at room temperature.
c)Chemical Name of the Element: Manganese
Chemical Symbol:Mg
Group it belong in periodic table:Mn
Other Element in the same group:Bohrium(Bh) ,Technetium(Tc)
Which represents the size of a population?
the birth rate
number of species in an area
the death rate
number of indivduals in a species
Pls hurry
Answer:
The Answer is D
Explanation:
Populations are shown by seeing how many speecies are in the area.
The number of individuals in a species represents the size of a population. hence, the last option is correct.
What is the population size?It is the actual no of individuals in the population.
The density of the population refers to the measurement of population size for per unit area.
So based on this, we can say that The number of individuals in a species represents the size of a population. hence, the last option is correct.
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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 525 mL of a solution that has a concentration of Na ions of 1.10 M
Answer:
31.652g of Na3PO4
Explanation:
We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:
Na3PO4 will dessicate in solution as follow:
Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)
From the balanced equation above,
1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.
Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e
xM Na3PO4 = (1.10 x 1)/3
xM Na3PO4 = 0.367M
Therefore, the molarity of Na3PO4 is 0.367M.
Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:
Molarity of Na3PO4 = 0.367M
Volume = 525mL = 525/1000 = 0.525L
Mole of Na3PO4 =..?
Molarity = mole /Volume
0.367 = mole /0.525
Cross multiply
Mole of Na3PO4 = 0.367 x 0.525
Mole of Na3PO4 = 0.193 mole.
Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:
Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol
Mole of Na3PO4 = 0.193 mole
Mass of Na3PO4 =.?
Mass = mole x molar mass
Mass of Na3PO4 = 0.193 x 164
Mass of Na3PO4 = 31.652g
Therefore, 31.652g of Na3PO4 is needed to prepare the solution.
A scientist measures the standard enthalpy change for the following reaction to be -115.5 kJ: CO(g) + Cl2(g)___COCl2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of COCl2(g) is ________ kJ/mol.
Answer:
-226.0kJ = ΔH°f COCl₂(g)
Explanation:
Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.
For the reaction:
CO(g) + Cl₂(g) → COCl₂(g)
Hess's law is:
ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))
ΔH°f CO(g) is -110.5kJ/mol
ΔH°f Cl₂(g) is 0 kJ/mol
Replacing in Hess's law:
-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)
-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ
-226.0kJ = ΔH°f COCl₂(g)a) What is the common-ion effect?b) Given an example of a salt that can decrease the ionization of HNO2 in solution. List all of the substances (species) that would be present in the solution. (Consider the HNO2, the salt, and the water.)
Answer:
Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.
NaNO2
Explanation:
In order to understand exactly what common ion effect is, let us consider a simple unambiguous example. Assuming I have a solution of an ionic substance that contains a cation A and an anion B, this ionic substance has chemical formula AB. Secondly, I have another ionic distance with cation C and anion B, its chemical formula is CB. Both CB and AB are soluble in water to a certain degree as shown by their respective KSp.
If I dissolve AB in water and form a solution, subsequently, I add solid CB to this solution, the solubility of CB in this solution is found to be lees than the solubility of CB in pure water because of the ion B^- which is common to both substances in solution. We refer to the phenomenon described above as common ion effect.
Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.
If I try to dissolve NaNO2 in a solution of HNO2, the solubility of NaNO2 in the HNO2 solution will be less than its solubility in pure water due to common ion effect. Also, the extent of ionization of HNO2 in a system that already contains NaNO2 will be decreased compared to its extent ionization in pure water. This system described here will contain HNO2, water and NaNO2
Combustion of hydrocarbons such as nonane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid nonane into gaseous carbon dioxide and gaseous water. 2. Suppose of nonane are burned in air at a pressure of exactly and a temperature of . Calculate the volume of carbon dioxide gas that is produced. Round your answer to significant digits.
Answer:
Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)
Part 2: Volume of CO₂ produced = 1223.21 L
Note: the complete second part of the question is given below:
2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.
Explanation:
Part 1: Balanced chemical equation
C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)
Part 2: volume of carbon dioxide produced
From the equation of the reaction;
At s.t.p., I mole of C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂
molar mass of C₉H₂₀ = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L
Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.
O.470 Kg of nonane = 470 g of nonane
470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂
Volume of CO₂ gas produced at 1 atm and 17 °C;
Using P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/P₂T₁
where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K
Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)
Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂
Calculate the temperature and state the appropriate phase(s) (solid, liquid, vapor) for each substance: This temperature is: Fahrenheit A. SubstanceMelting Point (K)Boiling Point (K)Phase(s) Oxygen, O254.7590.19 Methane, CH493.15109.10 Water, H2O273.15373.15
Answer:
Explanation:
At 54.75K melting point, Oxygen is in gas (vapour) phase
At 373.15K boiling point, water is in liquid phase.
At 109.10K boiling point methane is in gas (vapour) phase.
In a zero order reaction, it takes 342 seconds for 75% of a hypothetical reactant to decompose. Determine the half-life t_{1/2} in units of seconds. Do not enter units with your numerical answer. Numeric Answer:
Answer:
228 s
Explanation:
In a zero order reaction, the formula for the half life is given as;
t1/2 = [A]o / 2k
To obtain the rate constant k, we have to use;
[A] = [A]o - kt
kt = [A]o - [A]
From the question;
it takes 342 seconds for 75% of a hypothetical reactant to decompose.
We have;
t = 324
[A] = 25
[A]o = 100
Upon solving for k we have;
kt = [A]o - [A]
k = ( [A]o - [A] ) / t
k = (100 - 25 ) / 342
k = 75 / 342 = 0.2193
Solving for t1/2;
t1/2 = [A]o / 2k
t1/2 = 100 / 2(0.2193)
t1/2 = 100 / 0.4386 = 228 s
In supersonic flights, molecules break apart and react chemically. which safety features protect the plane?
Answer:
Explanation:
The heat Shield are materials (usually made of metals) protect us from heat by absoring lots of heat and gradually releasing heat by surrounding air cirucaltion
Answer:
Heat shield
Explanation: Most heat shields consist of one or more layers of stamped metal that are shaped into a shield that is designed to wrap around the exhaust manifold. The shield acts as a barrier and heat sink, preventing the heat from the manifold from reaching any of the components under the hood and potentially causing damage.
In a chemical equation, which symbol should be used to indicate that a substance is in solution? (s)
Answer:
(aq) meaning aqueous solution
Explanation:
hope it helps .
Give the expressions for the disappearance of the reactants first, in the order written in the chemical equation. Then write the expressions for the appearance of the products in the order written in the chemical equation.
Write the expressions in order of appearance in the equation in the form.
± 1/X . Δ[A]/ Δt
where ± is either a plus OR a minus sign, not both,
X is an integer, and
A is a chemical species.
4NH3(g)+ 5O2(g) → 4NO2+ 6H2O(g)
Answer: Rate of disappearance of [tex]NH_3[/tex] = [tex]\frac{-\Delta [NH_3]}{4dt}][/tex]
Rate of disappearance of [tex]O_2[/tex] = [tex]\frac{-\Delta [O_2]}{5dt}[/tex]
Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{+\Delta [NO_2]}{4dt}[/tex]
Rate of appearance of [tex]H_2O[/tex] = [tex]\frac{+\Delta [H_2O]}{6dt}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
For the given chemical equation :
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)[/tex]
Rate of disappearance of [tex]NH_3[/tex] = [tex]\frac{-\Delta [NH_3]}{4dt}][/tex]
Rate of disappearance of [tex]O_2[/tex] = [tex]\frac{-\Delta [O_2]}{5dt}[/tex]
Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{+\Delta [NO_2]}{4dt}[/tex]
Rate of appearance of [tex]H_2O[/tex] = [tex]\frac{+\Delta [H_2O]}{6dt}[/tex]
Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass of 10.88 g is added and the volume of the water now reads 49.4 mL. What is the density of the metal in g/mL? Multiple Choice 10.9 1.8 0.250 1.6 0.541
Answer:
1.8g
Explanation:
Initial volume = 43.5ml
Final volume = 49.4ml
Mass = 10.88g
Density = ?
Volume = Final volume - initial volume
= 49.4 - 43.5
= 5.9ml
Density = Mass/volume
Density = 10.88/5.9
= 1.8g/ml
When 2,2-dimethylbutane is subjected to free-radical chlorination, ________ distinct monochlorinated products are possible and ________ of these contain asymmetric carbon atoms.
Answer:
See the explanation
Explanation:
In this case, we have to remember that in the monochlorination products we only have to add one "Cl" with this in mind, we can have several options.
a) 1-chloro-2,2-dimethylbutane
b) (R)-3-chloro-2,2-dimethylbutane
c) (S)-3-chloro-2,2-dimethylbutane
d) 1-chloro-3,3-dimethylbutane
Additionally, from these 4 molecules, we will have 2 enantiomers. (R)-3-chloro-2,2-dimethylbutane and (S)-3-chloro-2,2-dimethylbutane.
See figure 1
I hope it helps!
There four distinct monochlorinated products and two of them contain an asymmetric carbon atom.
Free radicals refers to chemical species that posses an odd number of electrons. An asymmetric carbon atom is a carbon atom that is bonded to four different groups.
When 2,2-dimethylbutane is subjected to free-radical chlorination, four distinct monochlorinated products are possible and two of these contain asymmetric carbon atoms.
The four distinct monochlorinated products are;
1-chloro-2,2-dimethylbutane (R)-3-chloro-2,2-dimethylbutane (S)-3-chloro-2,2-dimethylbutane 1-chloro-3,3-dimethylbutane
Learn more: https://brainly.com/question/9743981
How many atoms are in 16.1 G Sr
Answer: There are [tex]1.11\times 10^{23}[/tex] atoms of Sr
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{16.1g}{87.6g/mol}=0.184moles[/tex]
1 mole of Sr contains = [tex]6.023\times 10^{23}[/tex] atoms of Sr
Thus 0.184 moles of Sr contains = [tex]\frac{6.023\times 10^{23}}{1}\times 0.184=1.11\times 10^{23}[/tex] atoms of Sr
Thus there are [tex]1.11\times 10^{23}[/tex] atoms of Sr
how to use VBT to draw the orbital overlapping diagram
Answer:
from the valence elecrtons configuration is the centre atom.atomised the number of elecrtons pair determine the hybridization .
Explanation:
you can read this note to know the ans