Answer:
See Explanations
Explanation:
pH =-log[H₃O⁺] = -log[H⁺]
pOH = -log[OH⁻]
For weak acids [H⁺] = SqrRt(Ka·[Acid])
For weak bases [OH⁻] = SqrRt(Kb·[Base])
pH + pOH = 14
__________________________________________
A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95
CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;
[OH⁻] = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M
=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05
=> pH = 14 - pOH = 14 - 2.05 = 11.95.
*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.
___________________________________________________
B. Given 0.18M CH₃NH₃Cl
In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴
Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.
Hydrolysis Reaction of Methylammonium Ion:
CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻
Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹ Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.
*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.
C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl
Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl
In Water ...
=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl
=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻
=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻
-----------------------------------------------------------
Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)
=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹
----------------------------------------------------------
From the 0.36M CH₃NH₃⁺
=> CH₃NH₃⁺ + H₂O ⇄ CH₃NH₄⁺ + OH⁻
C(eq) 0.36M ---- x x (<= at equilibrium after mixing)
Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)
=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M
=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process
=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻
--------------------------------------------------------
Cl⁻ + H₂O => No Rxn (Cl⁻ does not react with H₂O)The 0.1926M in OH⁻ => [H⁺] = Kw/[OH⁻] = (10⁻¹⁴/0.1926)M = 5.192 x 10⁻¹⁴M in H₃O⁺ ions (= H⁺ ions) ...∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix
The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and
CH₃NH₃Cl and the mixture give the following approximate values;
A. The pH value of the 0.18 M CH₃NH₂ is 11.93
B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69
C. The pH value of the mixture is 10.644
Which method can be used to calculate the pH values?A. 0.18 M CH₃NH₂
The solution is presented as follows;
CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻
Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we
have;
The number of moles of CH₃NH₂ remaining = 0.18 - x
Which gives;
[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]
[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴
Therefore;
[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]
4.167 × 10⁻⁴ × (0.18 - x) = x²
4.167 × 10⁻⁴ × (0.18 - x) - x² = 0
Which gives;
x = [OH⁻] = 8.455 × 10⁻³
pH = 14 + log[OH⁻]
Which gives;
pH = 14 + log(8.455 × 10⁻³) ≈ 11.93
B. 0.18 M CH₃NH₃Cl
The solution is presented as follows;
CH₃NH₃⁺ → CH₃NH₂ + H⁺
Let x represent the number of moles of CH₃NH₂ and H⁺ produced,
respectively, we have;
The number of moles of CH₃NH₃⁺ remaining = 0.18 - x
Which gives;
[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]
Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹
Therefore;
[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]
2.27 × 10⁻¹¹ × (0.18 - x) = x²
2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0
Which gives;
x = [H⁺] ≈ 2.02 × 10⁻⁶
pH = -log[H⁺]
Which gives;
pH = -log(2.02 × 10⁻⁶) ≈ 5.69
C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;
Based on the Henderson-Hasselbalch equation, we have;
[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]
Which gives;
[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]
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_________one moleculetwo moleculesone moletwo moles of H2(g) reacts with _________one moleculetwo moleculesone moletwo moles of F2(g) to form _________one moleculetwo moleculesone moletwo moles of HF(g).
Answer:
The reaction between hydrogen and fluorine.
Explanation:
The balanced chemical equation of the reaction is:
[tex]H_2(g)+F_2(g)->2HF(g)[/tex]
From the balanced chemical equation, it is clear that:
1 mole of [tex]H_2[/tex] gas reacts with one mole of [tex]F_2(g)[/tex] and forms 2 moles of HF(g).
Newly formed atoms, more neutrons, and kinetic energy are all:
A. necessary for nuclear fission reactions to occur.
B. sources of energy in a nuclear fission reaction.
C. products of nuclear fission reactions.
D. released when atoms decay.
Answer:
The correct answer is - C. products of nuclear fission reactions.
Explanation:
Nuclear fission is one of the nuclear reactions in which a heavy nucleus of the atom splits due to the impact of another particle or substance or it is on its own. In this reaction, there is a huge amount of kinetic energy released with more neutrons as the nucleus splits neutrons produced with energy.
The products of this reaction is newly formed atoms, kinetic energy and more neutrons and reactants is the heavy nucleus of the atom used.
it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to remove a single electron from n atom at the surface of the metal
Answer:
The right solution is "[tex]8.5\times 10^{-19} \ joule[/tex]".
Explanation:
As we know,
1 mole electron = [tex]6.023\times 10^{23} \ no. \ of \ electrons[/tex]
Total energy = [tex]513 \ KJ[/tex]
= [tex]513\times 1000 \ joule[/tex]
For single electron,
The amount of energy will be:
= [tex]\frac{513\times 1000}{(6.023\times 10^{23})}[/tex]
= [tex]8.5\times 10^{-19} \ joule[/tex]
A tank contains 100 grams of a substance dissolved in a large amount of water. The tank is filtered in such a way that water drains from the tank, leaving the substance behind in the tank. Consider the volume of the dissolved substance to be negligible. At what rate is the concentration (grams/liter) of the substance changing with respect to time in each scenario
Complete Question
A tank contains 100 grams of a substance dissolved in a large amount of water. The tank is filtered in such a way that water drains from the tank, leaving the substance behind in the tank. Consider the volume of the dissolved substance to be negligible. At what rate is the concentration (grams/liter) of the substance changing with respect to time in each scenario? (a) the rate after 5 hours, if the tank contains 60 L of water initially, and drains at a constant rate of 4 L/hr?
(b) the rate at the instant when 20 liters remain, if the water is draining at 2.4 L/hr at that instant g/L
c) the rate in scenario (b), if the unknown substance is also being added at a rate of 30 g/hr (and there are 100 grams in the tank at that instant)
Answer:
a) [tex]\frac{dc}{dt}=0.10g/l/hr[/tex]
b) [tex]\frac{dc}{dt}'=0.6g/l/hr[/tex]
c) [tex]\frac{dc}{dt}''=2.1g/l/hr[/tex]
Explanation:
From the question we are told that:
Mass m=100g
a)
Rate [tex]t=5[/tex]
Volume [tex]V=60L[/tex]
[tex]Drain Rate =4L/hr[/tex]
Generally the equation for Concentration is mathematically given by
[tex]c=\frac{mass}{Volume}[/tex]
Where
V=initial volume -(Drain Rate*time)
[tex]V=60-4t[/tex]
Therefore
[tex]c=\frac{100}{60-2.4t}[/tex]
[tex]\frac{dc}{dt}=\frac{240}{(60-4t)^2}[/tex]
At t=5
[tex]\frac{dc}{dt}=\frac{240}{(60-4(5))^2}[/tex]
[tex]\frac{dc}{dt}=0.250g/l/hr[/tex]
b)
Volume [tex]v=20[/tex]
Rate [tex]\frac{dv}{dt}= 2.4[/tex]
Generally the equation for Concentration is mathematically given by
[tex]c=\frac{100}{v}[/tex]
[tex]\frac{dc}{dt}=\frac{-100}{V_2^2}\frac{dv}{dt}[/tex]
[tex]\frac{dc}{dt}=\frac{-100}{20^2}*2.4[/tex]
[tex]\frac{dc}{dt}'=0.6g/l/hr[/tex]
c)
Rate [tex]\frac{dm}{dt}[/tex]
Generally the equation for Concentration is mathematically given by
[tex]c=\frac{m}{v}[/tex]
[tex]\frac{dc}{dt}=\frac{30*20-(-2.4(100))}{20L}[/tex]
[tex]\frac{dc}{dt}=\frac{30*20-(-2.4)(100)}{20}[/tex]
[tex]\frac{dc}{dt}''=2.1g/l/hr[/tex]
4. A salvage operator recovered coins believed to be gold. A sample weighed 385.000g and has a volume of 20.0mL. Were the coins gold (Density= 19.3g/mL) or yellow brass (Density =8.47g/mL)? Show your calculation and explain your answer.
Answer:
Gold
Explanation:
We are given that
Mass of sample ,m=385 g
Volume ,V=20mL
We have to find the coin is gold or yellow brass.
We know that
Density=[tex]\frac{Mass}{volume}[/tex]
Using the formula
[tex]Density=\frac{385}{20}g/mL[/tex]
[tex]Density=19.25g/mL[/tex]
[tex]Density\approx 19.3g/mL[/tex]
Density of coin=19.3g/mL
Density of gold=19.3g/mL
Hence, the coin is gold.
Define the following type of bonds..
Ionic bond.
Covalent bond.
Co ordinate bond separately .
Answer:
Ionic bond: Ionic bonds are bonds involving the attraction between oppositely-charged ions.
Covalent bond: Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms.
Co ordinate bond separately: a covalent bond in which both electrons are provided by one of the atoms. dative bond.
Explanation:
Hope this helps!
Nitrogen monoxide and water react to form ammonia and oxygen, like this: (g)(g)(g)(g) Write the pressure equilibrium constant expression for this reaction.
Answer:
Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}
Explanation:
First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:
NO(g) + H₂O(g) → NH₃(g) + O₂(g)
To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):
4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)
All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:
[tex]Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}[/tex]
dentify the correct formula for the following ionic compounds. - sodium chloride - magnesium chloride - calcium oxide - lithium phosphide - aluminum sulfide - calcium nitride A. SCl B. LiP 3 C. AlS D. Li 3P E. CaN F. CaO G. Ca 3N 2 H. MgCl 2 I. NaCl J. CaO 2 K. CaN 2 L. LiP M. MnCl 2 N. Al 2S 3 O. AlS 3
Explanation:
The chemical formula of an ionic compound can be written by using the symbols of the respective cations and anions.
The overall charge on the molecule should be zero.
Hence, the total charge of cations=total charge of anions.
The symbols of the given molecules are shown below:
sodium chloride ---- NaCl
magnesium chloride ---[tex]MgCl_2[/tex]
calcium oxide ---- CaO
lithium phosphide----[tex]Li_3P[/tex]
aluminum sulfide ----- [tex]Al_2S_3[/tex]
calcium nitride---- [tex]Ca_3N_2[/tex]
The rate constant of a first order reaction is 0.035 sec-1. How much of a 1.5 M sample of reactant would remain after 28 seconds
Answer:
0.56 M
Explanation:
Step 1: Given data
Rate constant (k): 0.035 s⁻¹Initial concentration of the reactant ([A]₀): 1.5 MTime elapsed (t): 28 sStep 2: Calculate the amount of reactant ([A]) after 28 seconds
For a first-order kinetics, we will use the following expression.
ln [A] = ln [A]₀ - k × t
ln [A] = ln 1.5 - 0.035 s⁻¹ × 28 s
[A] = 0.56 M
What is the major product in this reaction
Answer:
I think option A is right answer
for a reaction, ΔH= 206 kJ/mol and ΔS=0.215 kj/(k•mol). At what temperatures is this reaction spontaneous?
Answer:
When the temperature of the reaction is > 958.1K
Explanation:
A reaction is spontaneous when ΔG < 0.
We can solve this question finding the temperature at which ΔG < 0 using the equation:
ΔG = ΔH - TΔS
As ΔG < 0
ΔH < TΔS
206kJ/mol < T*0.215kJ/K.mol
958.1K < T
When the temperature of the reaction is > 958.1K, the reaction is spontanous because ΔG < 0.
2, classify the following molecules as polar or non polar.
A,CH4 B,CHcl C,Co2 D,H2O2 E,BCl3 F,H2S
A. CH4= NON POLAR
B. CH3cl= POLAR
C. CO2= NON POLAR
D. H2O2= POLAR
E. BCl3= NON POLAR
F. H2S= SLIGHTLY POLAR
Poly(ethylene terephthalate) (PET), which has glass transition (Tg) and crystalline melting (Tm) temperature of 69 and 267 °C, respectively, can exist in a number of different states depending upon temperature and thermal history. Thus, it is possible to prepare materials that are semicrystalline with amorphous regions that are either glassy or rubbery and amorphous materials that are glassy, rubbery or melts. Consider a sample of PET cooled rapidly from 300 °C (state A) to room temperature. The resulting material is rigid and perfectly transparent (state B). The sample is then heated to 100 °C and maintained at this temperature, during which time is gradually becomes translucent (state C). It is then cooled to room temperature, where it is again observed to be translucent (state D).
Answer:
Following are the solution to the given points:
Explanation:
For point A:
The sample cooking (PET) is between 300°C and room temperature.Now in nature, the substance is exceedingly stiff.Samples of PET up to 100°C were heated and stayed on equal footing.Now it has cooled off the same sample below 100° C and we may see how it is again TRASNEPARENT in nature.For point B:
In point 3, the mixture was added to 100°C, which implies that the granular material flows and deforms, enabling it to become elongated. This is termed solid-state crystalline such that grains are flexible, but this material contaminates numerous little crystalline that has spheres when we cool down in point 4 polymers. It forms therefore an unstructured solid, which then in point 4 is higher in particles and less pliable in orderly atoms.
For point C:
In point 2, the specimen gets forced at room temperature to organize a huge molecule in an ordinary and crystal fashion and therefore is transparent due to highly crystalline atoms in point 2 of the PET sample.
In point 4, however, we notice how amorphous, firm but not crystalline develops. It's why light tends to disperse over many cereal limits, since many microscopic crystallines, therefore dispersion, PET in point 4 is translucent.
cual es la masa atomica del hidrogeno
El hidrógeno es el elemento químico de número atómico 1, representado por el símbolo H. Con una masa atómica de 1.00784 u es el más ligero de la tabla periódica de los elementos. Por lo general, se presenta en su forma molecular, formando el gas diatómico H₂ en condiciones normales.
A chemical reaction was carried out by mixing 25 g of pure CaCO3 and 0.75 mole of pure HCl to give CaCl2, H2O and CO2. a. Which one is the limiting reactant and why? b. Calculate the mass of CaCl2 produced. c. How many number of water molecules are formed? d. Calculate the volume of CO2 gas liberated at STP. e. What mass of NaOH is required to absorb the whole CO2 produced in the reaction?
hola, esta pregunta es bastante difícil pero está bien, no lo sé, lo siento :) :)
All the options are solved and answer is written below
What is a Chemical Reaction ?A reaction between two or more compounds to form products made after chemical change is called a chemical reaction.
It is given that
A chemical reaction was carried out by mixing 25 g of pure CaCO₃ and 0.75 mole of pure HCl
CaCl₂ , H₂O and CO₂ are the products obtained.
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Mole ratio CaCO₃ : HCl : CaCl₂ : H₂O = 1 : 2 : 1 : 1
Molar mass of CaCO₃ = 100 g/mol
Molar mass of HCl = 36.5 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of CaCl₂= 110.98 g/mol
Moles of CaCO₃ = 25/100 = 0.25 moles
Moles of HCl present = 0.75 mole
For 0.25 moles of CaCO₃ 0.5 moles of HCl is required , as the moles of HCl is present in excess therefore
a. CaCO₃ is the limiting reactant
b. mass of CaCl₂ produced
Moles of CaCl₂ produced = 0.25 moles
1 mole means 110.98 gm
0.25 mole means 0.25* 110.98 = 27.74 gm
c. moles of water molecules formed
for 0.25 moles of CaCO₃ 0.25 moles of water will be formed
d.Volume of Co produced at STP
PV = nRT
P= 1 atm
V=?
R = 0.0821 atm L/K/mol
V = 0.25 * 0.0821 * 273 /1
V = 5.6 liter
e. The mass of NaOH required to absorb CO₂ produced in the reaction
Ratio of NaOH:CO₂ = 2 :1
0.5 moles will be required , i.e.
0.5 *40
20 grams of NaOH will be required.
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what is the ph of a 0.005 M aq. HClsolution
Answer:
Explanation:
since HCL is a strong acid, -log(0.005) = 2.30 pH
A sample of oxygen gas has a volume of 89.6 L at STP. How many moles of oxygen gas are present ?
Answer:
89,6/22,4 =4(mol)
Explanation:
There are approximately 1.089 moles of oxygen gas present in the sample at STP.
At STP (Standard Temperature and Pressure), the conditions are defined as follows:
Temperature (T) = 0 degrees Celsius = 273.15 Kelvin
Pressure (P) = 1 atmosphere (atm) = 101.325 kPa = 1013.25 hPa
Now, to find the number of moles of oxygen gas (O2) present in the sample, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant = 0.0821 L.atm/(mol.K)
T = temperature (in Kelvin)
Given:
V = 89.6 L (volume at STP)
T = 273.15 K (STP temperature)
Let's plug in the values and solve for n (number of moles):
n = PV / RT
n = (1 atm) × (89.6 L) / (0.0821 L.atm/(mol.K) × 273.15 K)
n = 1.089 moles
So, there are approximately 1.089 moles of oxygen gas present in the sample at STP.
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Balance chemical equations
Answer:
you first get the unbalanced equation to show the various products and reactants. then, you write down the atoms in each element(quantity). add the coefficients so it's equal on both sides, and get state of matter
Explanation:
What does a positive AH tell about a reaction?
A. The reaction is exothermic.
B. The reaction has heat as a product.
C. The reaction is endothermic.
D. The reaction has no activation energy.
Answer:
C
Explanation:
An endothermic reaction has a positive enthalpy change (∆H> 0).
An endothermic reaction is where the energy of the products is higher than that of the reactants.
∆H= energy of products -energy of reactants
Thus, ∆H is positive since the value of the energy of products is greater than that of the reactants.
Exothermic reactions have a negative ∆H.
In an endothermic reaction, heat is absorbed and thus if we were to include heat in teh chemical equation, it would be part of the reactants not products.
∆H does not indicate the amount of activation energy (Ea). All reactions have activation energy (exothermic and endothermic reactions). Activation energy is the minimum amount of energy required for the reaction to proceed.
Answer:
[tex]\boxed {\boxed {\sf C. \ The \ reaction \ is \ endothermic}}[/tex]
Explanation:
There are two main types of reactions classified according to heat: exothermic and endothermic.
Exothermic: heat is released from the system Endothermic: heat is absorbed into the systemThe ΔH is the change in enthalpy. It is the difference between the heat of the products and the reactants (ΔH = heat of products - heat of reactants). It helps us describe a system's change in heat and classify reactions as exothermic or endothermic.
Exothermic: the products have less energy because heat is released. The change in enthalpy or ΔH is negative.Endothermic: the products have more energy because heat is absorbed. The change in enthalpy or ΔH is positive.In this problem, the change in enthalpy is positive. The change in enthalpy doesn't refer to heat as a product or activation energy. Therefore, the reaction must be endothermic.
Base your answer(s) to the following question(s)
on the data table below and on your knowledge of
biology
A group of students obtained the following
data:
Data Table
Student Pulse Rate at Pulse Rate After
Tested Rest
Exercising
1
70
97
2
75
106
3
84
120
4
60
91
5
78
122
Answer:
122
Explanation:
because after exercises heart beat increases
How many electrons are shown in the following electron
configuration: 1s22s22p63s 23p64s23d104p65s24d105p66s2 ?
Express your answer numerically as an integer.
Answer:
1s22s22p6
Explanation:
Neon is an element in the periodic table and has an atomic number of 10, which means it has 10 protons in its nucleus and thus since the number of protons and electrons is the same then it has 10 electrons.
Therefore, it has 2 electrons in the first energy shell and 8 electrons in the second energy shell. To elaborate further, the first shell has a single s-sub shell that contains a single s-orbital that can hold two electrons. The second energy shell has a single s-sub-shell whose s-orbital will occupy 2 electrons, and also has a p-orbital which can hold 6 electrons, making the second shell to have 8 electrons.
Write balanced chemical equations for the following reactions:
a) hydrogen gas + nitrogen gas gives ammonia.
b) sodium peroxide + water gives sodium hydroxide + oxygen gas.
If you can explain how to do it it'll be greatly appreciated.
Answer:
a) 3H₂ (g) + N₂ (g) → 2NH₃ (g)
b) 2Na₂O₂ + 2H₂O → 4NaOH + O₂
General Formulas and Concepts:
Atomic Structure
Reading a Periodic TableReactions rxnCompoundsAqueous Solutions
States of matterExplanation:
a)
When we write this chemical reaction, we know that hydrogen and nitrogen are diatomic elements. Ammonia you just have to remember the chemical compound formula for. So our unbalanced equation would be:
H₂ (g) + N₂ (g) → NH₃ (g)
Now to balance this equation, we see that we have an uneven amount of hydrogens and nitrogens on both sides of the rxn. Let's balance out the nitrogens first by multiplying the products by 2:
H₂ (g) + N₂ (g) → 2NH₃ (g)
We see that now we have the number of nitrogens balanced on both sides, but our hydrogens are still unbalanced. Let's balance those by making the reactants the same number as our products:
We have 6 hydrogens now on the products side2H = 6HIt looks like we need to multiply 3 on the reactant hydrogens:
3H₂ (g) + N₂ (g) → 2NH₃ (g)
And we have our balanced formula!
b)
Same concept as A.
Recall how to write chemical compounds. The charge of sodium (Na) is +1 and the charge for polyatomic ion peroxide (O₂²⁻) is -2. Also recall the charge for polyatomic ion hydroxide (OH)m which is -1:
Sodium peroxide = Na₂O₂
Water = H₂O (standard knowledge)
Sodium hydroxide = NaOH
Oxygen gas = O₂
Write out our unbalanced rxn:
Na₂O₂ + H₂O → NaOH + O₂
Right away we can see that it is definitely unbalanced. We can see that we have an odd number of oxygens on both sides. We don't like odds here, so let's multiply 2 to the sodium peroxide and to make it even:
2Na₂O₂ + H₂O → NaOH + O₂
We can see that we have an even amount of oxygens on the reactant side. Now we have to balance the number of sodiums on the product side:
2Na₂O₂ + H₂O → 4NaOH + O₂
We now have the sodiums balanced. Moving onto the hydrogens. 2 on the reactant side and 4 on the product side:
2Na₂O₂ + 2H₂O → 4NaOH + O₂
We now have the hydrogens balanced. When we move on to oxygens, we can see that the number of oxygens have the same number of moles on both sides, and that would be our balanced rxn.
Answer:
Solution given:
a) hydrogen gas + nitrogen gas gives ammonia.
Balanced chemical equation:
[tex]\boxed{\bold{\green{3H_{2}+N_{2}\rightarrow 2NH_{3}}}}[/tex]
when pure and dry Nitrogen and Hydrogen gas is passed in the ratio 2:3 at 450°C temperature 200-900 ATM pressure in tye presence of iron as catalyst and molybdenum as promotor which forms Ammonia.
b) sodium peroxide + water gives sodium hydroxide + oxygen gas.
Balanced chemical equation:
[tex]\boxed{\green{\bold{3NaO_{2}+2H_{2}O \rightarrow 4NaOH +O_{2}}}}[/tex]
When sodium peroxide is combined with hot water double displacement reaction takes place and forms sodium hydroxide and oxygen gas along with heat.
What type of reaction?
Use the following key to classify each of the elements below in its elemental form:
A. Discrete atoms .. C. Metallic lattice
B. Molecules ... D. Extended, three-dimensional network
1. Magnesium
2. Nitrogen ...
3. Lithium
4. Potassium ...
Answer:
Magnesium - Metallic lattice
Nitrogen - Molecules
Lithium - Metallic lattice
Potassium - Metallic lattice
Explanation:
Metals exist in metallic lattices. In this lattice, metal ions are held together with a sea of electrons by strong electrostatic forces.
All metals possess this metallic lattice, hence; potassium, lithium and magnesium all consist of metal lattices.
Nitrogen is a nonmetal and consists of molecules of N2.
Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon.
The question is incomplete, the complete question is;
Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon
a) electronic excitation
b) bond breakage
c) molecular vibration
d) molecular rotation
Answer:
molecular rotation
Explanation:
Microwaves are part of the electromagnetic spectrum. They are lower energy, lower frequency radiation.
When molecules absorb infrared radiation, they transition between the rotational states of the molecule.
Hence, the highest energy molecular process that occurs when a molecule absorbs a microwave photon is molecular rotation.
6.50 g of a certain Compound x, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 128. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product carbon dioxide water mass 22.35 g 3.66 g Use this information to find the molecular formula of x
Answer:
[tex]C_{10}H_8[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to find the molecular formula of the given compound by firstly calculating both moles and grams of carbon in carbon dioxide and hydrogen in water, as the only sources of these elements derived from the compound x due to its combustion:
[tex]n_C=22.35gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.51molC\\\\m_C=0.51molC*\frac{12.01gC}{1molC} =6.10gC[/tex]
[tex]n_H=3.66gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.41molH\\\\m_H=0.41molH*\frac{1.01gH}{1molH}=0.41g[/tex]
Now, since the addition of carbon and hydrogen is about 6.50 grams, we infer the compound has no oxygen, that is why we now set the mole ratios in the empirical formula for both C and H as shown below:
[tex]C:\frac{0.51mol}{0.41mol}= 1.24\\\\H:\frac{0.51mol}{0.51mol}= 1\\\\C_{1.24}H[/tex]
Yet it cannot be decimal, that is why we multiply by 4 to get the correct whole-numbered empirical formula:
[tex]C_5H_4[/tex]
Whose molar mass is 64.09 g/mol, which makes the ratio of molar masses:
[tex]\frac{128.g/mol}{64.09g/mol} =2[/tex]
Therefore, the molecular formula is twice the empirical one:
[tex]C_{10}H_8[/tex]
Regards!
A gas at 74°C is heated to 120°C so there is pressure reaches 1.79 ATM. What is its initial pressure?
Explanation:
here's the answer to your question
A reaction at evolves of dinitrogen monoxide gas. Calculate the volume of dinitrogen monoxide gas that is collected. You can assume the pressure in the room is exactly . Be sure your answer has the correct number of significant digits.
Answer:
The answer is "[tex]V = 18.8\ L[/tex]".
Explanation:
Given:
[tex]n= 0.788 \ mmol = 0.788 \ mol \ \text{(1 mmol is a thousandths of a mol)}\\\\R = 0.08206\ \frac{Latm}{Kmol}\\\\T = 17.0^{\circ}\ C = (17.0 + 273) \ K = 290 \ K\\\\ p=1\ atm\\\\V = ?[/tex]
Using idel gas law:
[tex]PV = nRT\\\ \therefore \\\\V= \frac{nRT}{P}[/tex]
[tex]V =\frac{0.788 \times 0.08206 \times 290}{1}[/tex]
[tex]=\frac{18.7523}{1} \\\\\ = 18.75[/tex]
Which group of nonmetals is the most reactive?
A. Halogen
B. Alkali
c. Noble gases
D. Oxygen family
Answer:
Halogens
HalogensHalogens are the most reactive nonmetals on the periodic table.
Answer:
A) halogen
Explanation:
halogen are highly reactive because they are all electro negative
The reversible reaction: 2SO2(g) O2(g) darrow-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2
Answer:
[tex][O_2]_{eq}=0.030M[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the mathematical expression for the concentration of oxygen at equilibrium, given the initial one and the change due to the reaction extent:
[tex][O_2]_{eq}=0.050M-x[/tex]
Whereas [tex]x[/tex] can be found considering the equilibrium of SO3:
[tex][SO_3]_{eq}=2x=0.040M[/tex]
Which means:
[tex]x=\frac{0.040M}{2} =0.020M[/tex]
Thus, the equilibrium concentration of oxygen gas turns out:
[tex][O_2]_{eq}=0.050M-0.020M=0.030M[/tex]
Regards!
