Answer:
Pressure, P = 67.57 atm
Explanation:
Given the following data;
Volume = 0.245 LNumber of moles = 0.467 molesTemperature = 159°CIdeal gas constant, R = 0.08206 L·atm/mol·KConversion:
We would convert the value of the temperature in Celsius to Kelvin.
T = 273 + °C
T = 273 + 159
T = 432 Kelvin
To find the pressure of the gas, we would use the ideal gas law;
PV = nRT
Where;
P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.Making P the subject of formula, we have;
[tex] P = \frac {nRT}{V} [/tex]
Substituting into the formula, we have;
[tex] P = \frac {0.467*0.08206*432}{0.245} [/tex]
[tex] P = \frac {16.5551}{0.245} [/tex]
Pressure, P = 67.57 atm
Which of these are characteristics of good experimental design
Two substances (A and B) are placed at each end of a glass tube. The vapors of the substances form a solid that is closer to substance
A's end than to substance B's end.
Which conclusion will be made?
O Substance A has a greater molecular mass than substance B.
Substance B has a greater molecular mass than substance A.
The molecular masses of both substances are the same.
Answer:
Substance A has a greater molecular mass than substance B
Explanation:
The greater the mass of a substance, the slower it will travel. If the solid formed closer to substance A, it means that substance A, travelled less and that it was slower, meaning it had a greater molecular mass.
Please mark it as brainliest if it helped you
Answer:
thank you... I would say is the answer: Subtance A has a greater molecular mass than substance B.
Explanation:
thanks for posting all the comments all the time!
have a great week... stay warm from a snow storm
(1) Some years ago, a crucial experimental error was made during the work-up of the reaction. Instead of adding the reaction mixture to the cold acid solution, the inverse addition was carried out. What do you think happened when the acid solution was added to the reaction mixture
Answer:
Salts will be formed and the pH of the new mixture will fall.
Explanation:
What should be: The reaction mixture should be added to the cold acid solution.
The error: Cold acid solution was added to the reaction mixture.
What happens with this?
The neutralization of a solution creates salts. The combination of positive and negative ions in the reaction also forms water. Assuming the pH of the reaction mixture was neutral before the addition of the acid, after the addition of the acid solution the pH will drop. Recall that a neutral solution has a pH of 7. So the pH of this new solution will drop to less than 7 because of the addition of an acid.
If the required procedure for the chemical process was followed, the reaction mixture added to the cold acid would have resulted in a stronger base solution. In other words, the pH of the new solution would be higher.
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.
When a solution is saturated, more solute will typically dissolve if you: ___________. A. change the temperature B. stir the solution C. continue to add solute D. crush or grind the solute into a fine powder
Answer:
A
Explanation:
More solute would typically dissolve in a saturated solution if the temperature of the solution is raised.
A saturated solution is a solution that contains the maximum amount of a particular solute the solvent can dissolve at a particular temperature.
Stirring may only increase the rate at which a solute dissolves in a solvent to make a solution but will not make more solute to dissolve in a saturated solution.
Crushing or grinding solutes into a fine powder may also increase the rate at which such solutes dissolve in a solvent to make a solution. However, it will not increase the capacity of a saturated solution to dissolve more of the solute.
Solubility is generally known to increase with an increase in temperature because the increase in the kinetic energy of the molecules of the solvent results in an increase in the efficiency with which they break apart or dissolve the solute molecules.
The correct option is, therefore, A.
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
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:
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
Antacids are
A)
weak acids.
B)
weak bases.
C)
strong bases.
D)
strong acids.
Answer: B- Weak base.
Explanation:
Answer:
Weak bases is your answer
Explanation:
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
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]
Calculate the various ratios of 2-methylbutane
C₅H₁₂
Isopentane/Formula
The preferred IUPɑC nɑme is the systemɑtic nɑme 2-methylbutɑne. ɑn isopentyl group is ɑ subset of the generic pentyl group. It hɑs the chemicɑl structure -CH3CH2CH(CH3)2.
states two properties a solute need to satisfy to be responsible for the colligative properties?
Answer:
the properties are:
vapor pressure loweringosmotic pressurefreezing point depressionboiling point elevationthese are all the properties but I think the two a solute needs to satisfy are
boiling point elevationvapor pressure loweringI hope this helps
Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma bond. Compound A is a: cumulated diene conjugated diene isolated diene Compound B: Two alkenes are joined by a C H 2 group. Compound B is : isolated diene conjugated diene cumulated diene Compound C: Two alkenes are joined by C H 2 C H 2. Compound C is a: conjugated diene isolated diene cumulated diene Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an s p 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent. Compound D is a: isolated diene conjugated diene cumulated diene
Explanation:
Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.
Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.
Isolated double-bonded compound has a single bond isolated by two to three single bonds.
Compound A: Two alkenes are joined by a sigma bond.
For example:
[tex]-CH_2=CH-CH=CH2-[/tex]
It is a conjugated diene.
Compound B: Two alkenes are joined by a C H 2 group.
It is a cumulative diene.
Compound C: Two alkenes are joined by C H 2 C H 2.
Then it is an isolated alkene.
Compound D: A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.
Hence, compound D is a conjugated diene.
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.
What type of reaction?
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!
2. How many joules of heat are released when 32g of water cools down from 71%
specific heat of water is 4.184 J/gºC)
How many kilojoules is this?
he says he doesnt know sorry
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.
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 417 mL .
Answer:
m = 1125.9 g.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the definition of density as mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we solve for the mass in the equation to obtain:
[tex]m=d*V[/tex]
Then, we plug in the values to obtain:
[tex]m=2.70g/mL*417mL\\\\m=1125.9g[/tex]
Regards!
A rectangular piece of plastic has a width of 4.2 cm, a height of 1.9 cm and a length of 8.8 cm. If the mass of the plastic rectangle is 64.6 g, what is its density in g/mL?
Answer:
0.92g/mL
Explanation:
Density of a substance is calculated as follows:
Density = mass (m) ÷ volume (V)
According to this question, a rectangular piece of plastic has a width of 4.2 cm, a height of 1.9 cm and a length of 8.8 cm. Using the formula; L × W × H, the volume of the plastic can be calculated
V = L × W × H
V = 8.8 × 4.2 × 1.9
V = 70.2cm³
The mass of the plastic is 64.6g, hence, its density is:
Density = 64.6g ÷ 70.2cm³
Density of the rectangular plastic = 0.92g/cm³ or 0.92g/mL
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
What is true about the properties of liquids and gases?
Gas particles are much more densely packed than liquid particles.
The crystal lattice structure of liquids is more defined than in gases.
Liquids form amorphous crystals while gases do not.
There are strong intermolecular forces between particles that make up liquids, but not gases.
Answer:
There are strong intermolecular forces between particles that make up liquids, but not gases.
Explanation:
Solids, liquids and gases are the three states of matter that exists. However, they possess varying properties that distinguishes them from one another. One of these properties is the strength of the intermolecular forces that hold their molecules together.
The intermolecular forces of each state of matter becomes weak in this order: solid>liquid>gas.
- Intermolecular forces in solid molecules are very strong, hence making them compact and well attached to each other.
- Intermolecular forces in liquid molecules are not too strong, hence, cannot exist in a fixed position but tend to flow.
- Intermolecular forces in gaseous molecules are very weak, hence, gases can move easily and rapidly in any given space.
complete the following steps.
Remember to follow lower numbered rules first.
Na2CO3(aq) + Pb(OH)2(aq) → NaOH (?) + PbCO3(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write "no
reaction.” (1 pt)
Answer:
See explanation
Explanation:
a) The balanced reaction equation is;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3
b) When we include states of matter;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)
c) Complete ionic equation;
2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)
Net Ionic equation;
Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)
Help!!!!!!!!!
I'm using plato
Answer:
- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.
- One black ball and two black balls: they represent a compound formed by two different elements.
- One gray ball and two black balls: they represent a compound formed by two different elements.
- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.
Explanation:
Hey there!
In this case, according to the given information, we can firstly bear to mind the fact that each ball color represents a different element, for that reason we can tell the following:
- Two black balls: they represent a diatomic molecule composed by two atoms of the same element.
- One black ball and two black balls: they represent a compound formed by two different elements.
- One gray ball and two black balls: they represent a compound formed by two different elements.
- Two black-dotted balls: they represent a diatomic molecule composed by two atoms of the same element.
Regards!
A container is filled to a volume of 55.2 L at 61 °C. While keeping the
temperature constant, the volume is reduced to 28.8 L and the pressure at
the end was recorded to be 8.53 atm. What was the initial pressure inside
the container, in units of atm?
Answer:
4.45 atm
Explanation:
Applying,
PV = P'V'............ Equation 1
Where P = Initial pressure of the container, V = Initial volume of the container, P' = Final pressure of the container, V' = Final volume of the container.
make P the subject of the equation
P = P'V'/V........... Equation 2
From the question,
Given: V = 55.2 L, P' = 8.53 atm, V' = 28.8 L
Substitute these values into equation 2
P = (8.53×28.8)/55.2
P = 4.45 atm
Answer:
[tex]\boxed {\boxed {\sf 4.45 \ atmospheres}}[/tex]
Explanation:
We are asked to find the pressure given a change in volume. The temperature remains constant, so we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:
[tex]P_1 V_1= P_2V_2[/tex]
The initial pressure is unknown, but the volume starts at 55.2 liters.
[tex]P_1 * 55.2 \ L = P_2V_2[/tex]
The volume is reduced to 28.8 liters and the pressure is 8.53 atmospheres.
[tex]P_1 * 55.2 \ L = 8.53 \ atm * 28.8 \ L[/tex]
We are solving for the initial pressure, so we must isolate the variable P₁. It is being multiplied by 55.2 liters. The inverse operation of multiplication is division, so we divide both sides of the equation by 55.2 L.
[tex]\frac {P_1 * 55.2 \ L }{55.2 \ L}= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}[/tex]
[tex]P_1= \frac{8.53 \ atm * 28.8 \ L}{55.2 \ L}[/tex]
The units of liters (L) cancel.
[tex]P_1= \frac{8.53 \ atm * 28.8 }{55.2}[/tex]
[tex]P_1=\frac{245.664 }{55.2 } \ atm[/tex]
[tex]P_1 = 4.45043478261 \ atm[/tex]
The original measurements of volume and pressure have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 0 in the thousandths place tells us to leave the 5.
[tex]P_1 \approx 4.45 \ atm[/tex]
The initial pressure inside the container is approximately 4.45 atmospheres.
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.
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.
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.
In an experiment 25.0 mL of 0.100 M KI was diluted to 50.0 mL. Calculate the molarity of the diluted solution
Answer:
The molarity is "0.050 M".
Explanation:
The given values are:
M1 = 0.100 M
M2 = ?
V1 = 25.0 mL
V2 = 50.0 mL
As we know,
⇒ [tex]M1\times V1=M2\times V2[/tex]
Or,
⇒ [tex]M2=\frac{M1\times V1}{V2}[/tex]
By putting the values, we get
[tex]=\frac{0.100\times 25}{50}[/tex]
[tex]=\frac{2.5}{50}[/tex]
[tex]=0.05 \ M[/tex]
