Answer:
The energy of particles of matter determines the matter's state. Particles of a gas have more energy than particles of a liquid, and particles of a liquid have more energy than particles of a solid. Therefore, in order for matter to change from a solid to a liquid or from a liquid to a gas, particles of matter must absorb energy. In order for matter to change from a gas to a liquid or from a liquid to a solid, particles of matter must lose energy.
The system will lose or gain energy as a result of changing its state, which is why energy will be involved in state changes.
What is energy?
Energy is a quantitative property that would be transferred to a body or even a physical system and is visible in term of heat as well as light during the performance of work.
Whenever a state changes, energy is involved. When matter transitions through one state to another, it either releases or absorbs energy. When matter transforms from such a liquid to a solid, for example, it loses energy. When matter transforms from a solid to a liquid, the opposite occurs.
To know more about energy.
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What is the mass in grams of 2.5 moles of Al?
Answer:
One mole of Al weighs 27g.
2.5 moles of Al weigh 67.5g.
How long would it take a bus traveling 52 km/h to travel 130 km
Answer:
2 and a half hours
Explanation:
Time is equal to distance over speed
3. How can the force of gravity affect weight?
An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant. B) an increase in reaction rate due to a decrease in the kinetic energy of the reactants. C) an increase in the rate of reaction because reactant molecules collide with greater energy. D) an increase in both the reaction rate and activation energy due to increased kinetic energy.
Answer:
C) an increase in rate of reaction because reactant molecules collide with greater energy
Explanation:
Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.
The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.
A geochemist has determined by measurements that there are moles of tin in a sample of cassiterite. How many moles of oxygen are in the sample
The question is incomplete, the complete question is:
This is the chemical formula for cassiterite (tin ore): SnO2 A geochemist has determined by measurements that there are 13. moles of tin in a sample of cassiterite. How many moles of oxygen are in the sample? Round your answer to 2 significant digits.
Answer:
Explanation:
In SnO2, there are two moles of oxygen for each mole of tin.
Hence, if there are 13 moles of tin, then we should have 13 * 2 moles of oxygen. This gives us 26 moles of oxygen.
Hence there are 26 moles of oxygen.
9th grade science need help ASAP
Answer:
D
Explanation:
It is a solid everything that is a solid is heavier
Which of the items below is a colloid?
a.fruit salad b.gelatin c.lacquer
Answer:
b. gelatin
Explanation:
a homogeneous noncrystalline consisting of large molecules or ultramicroscopic particles of one substance.
What does the term mass mean in science?
A. The measure of the force of gravity on an object.
B. The amount of matter in an object.
C. The Amount of space that matter takes up.
Answer:
B
Explanation:
Mass is defined as the quantity of matter in an object.
If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction
Answer:
7.71 × 10⁻⁴ M/s
Explanation:
The initial rate of the reaction can be expressed by using the formula:
[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]
where;
Pressue P = 1.00 atm
Volume V =5.74mL = (5.74 /1000) L
Rate R = 0.082 L atm/mol.K
Temperature = 298 K
[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]
= 2.35 × 10⁻⁴ mol
Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]
Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]
Δ[O₂] = 0.04626 M
The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
= [tex]\dfrac{0.04626}{60}[/tex]
= 7.71 × 10⁻⁴ M/s
The ion with the smallest diameter is ________. The ion with the smallest diameter is ________. Be2 Sr2 Ca2 Ba2 Mg2
Answer:
Be2^+
Explanation:
Ionic diameter increases down the group. This implies that Be2^+ will have the smallest diameter.
This extremely small diameter makes Be2^+ to differ considerably from other ions of group 2 elements.
For instance, the compounds of beryllium are mostly covalent in nature.
Gases A and B are confined to a cylinder and piston and react to form product C. As the reaction occurs, the system loses 1189 J of heat to surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 311 J of work on the system. What is the change in the internal energy of the system
Answer:
The change in the internal energy of the system -878 J
Explanation:
Given;
energy lost by the system due to heat, Q = -1189 J (negative because energy was lost by the system)
Work done on the system, W = -311 J (negative because work was done on the system)
change in internal energy of the system, Δ U = ?
First law of thermodynamics states that the change in internal energy of a system (ΔU) equals the net heat transfer into the system (Q) minus the net work done by the system (W).
ΔU = Q - W
ΔU = -1189 - (-311)
ΔU = -1189 + 311
ΔU = -878 J
Therefore, the change in the internal energy of the system -878 J
which material should you use so that the area is cool in terms of temperture O asphalt O red bricks O concrete O soil O
Answer:
red bricks
Explanation:
right on edg 2020
Red bricks should you use so that the area is cool in terms of temperture. Therefore, option B is correct.
What are red bricks used for?Red bricks can be used in the construction of structures such as buildings, foundations, arches, pavement, and bridges. These can also be used for aesthetic purposes such as landscaping, face work, and a variety of other architectural purposes.
Traditional red bricks are saw to be more robust, and structures constructed with them are stronger than hollow block structures.Red brick can be warm, with rust or terracotta undertones. They could also be cool and more burgundy.
Thus, Red bricks should you use so that the area is cool in terms of temperture, option B is correct.
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what kind of bonds are there in H2O?
Answer: it is covalent and there are 2 hydrogen molecules and 1 oxygen molecule.
Explanation: it just is
Rahul and Manav each were given a mixture of iron filings and sulphur powder. Rahul heated the mixture strongly and a new substance was formed. Write three points of difference between the two.Required to answer.
Answer:
sure
Explanation:
The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.
The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.
The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.
1.
Light near the middle of the ultraviolet region of the electromagnetic spectrum
has a frequency of 2.73 X 1016 s-1
a.
What is the wavelength of this radiation in meters (m)?
b.
What is the energy associated with this radiation in kcal?
Explanation:
Given that,
The frequency of electromagnetic spectrum is [tex]2.73\times 10^{16}\ Hz[/tex]
(A) Let the wavelength of this radiation is [tex]\lambda[/tex]. We know that,
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{2.73\times 10^{16}}\\\\\lambda=1.09\times 10^{-8}\ m[/tex]
So, the wavelength of this radiation is [tex]1.09\times 10^{-8}\ m[/tex].
(B) Let E is the energy associated with this radiation. Energy of an electromagnetic radiation is given by :
[tex]E=hf[/tex]
h is Planck's constant
[tex]E=6.63\times 10^{-34}\times 2.73\times 10^{16}\\\\E=1.8\times 10^{-17}\ J[/tex]
1 kcal = 4184 J
It means,
[tex]1.8\times 10^{-17}\ J=\dfrac{1}{4184}\times 1.8\times 10^{-17}\\\\=4.3\times 10^{-21}\ \text{kcal}[/tex]
Hence, this is the required solution.
On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential energy of two O atoms versus the distance between their nuclei.
Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
For the combustion of methane presented in Example 5.4, the chemical reaction is CH4 +2O2 →CO2 +2H2O Suppose that methane flows into a burner at 30 gmol/s, while oxygen flows into the same burner at 75 gmol/s. If all the meth- ane is burned and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? Hint 1: Does the fact that all the methane is burned mean that all the oxygen is burned also? Hint 2: Find the molar flow rate of each component gas in the outlet gas ("flue gas").
Answer:
[tex]x_{CO_2}^{out} =0.25[/tex]
Explanation:
Hello.
In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:
[tex]CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon[/tex]
Whereas [tex]\epsilon[/tex] accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:
[tex]0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s[/tex]
Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:
[tex]f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s[/tex]
It means that the mole fraction of carbon dioxide in that output is:
[tex]x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25[/tex]
Best regards.
How many carbon atoms are in vitamin c?
Answer:
molecules can be much bigger. one molecule of vitamin c is made up of 20 atoms (6 carbons, 8 hydrogens, and 6 oxygens
What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.
Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
The first temperatures is [tex]T_1 = 25^oC = 25 +273 =298 \ K[/tex]
The second temperature is [tex]T_2 = 100^oC = 100 +273 = 373 \ K[/tex]
Generally the equation for the most highly populated rotational energy level is mathematically represented as
[tex]J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also
B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
Generally the energy require per mole to move 1 cm is 12 J /mole
So [tex]0.244 \ cm^{-1}[/tex] will require x J/mole
[tex]x = 0.244 * 12[/tex]
=> [tex]x = 2.928 \ J/mol [/tex]
So at the first temperature
[tex]J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 20 [/tex]
So at the second temperature
[tex]J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 22.5 [/tex]
The wood in my house is crumbling. *
Problem
Hypothesis
Law
Theory
Answer:
Problem
Explanation:
The given statement is a problem. It states the problem that the house of the speaker has been undergoing with. This problem gives rise to the Hypothesis in which the 'why' question is asked. The reason of the crumbling of the wood is stated in the hypothesis. Any reason placed of the happening of the event is stated to be hypothesis.
which angles are right
Answer:
a right is 90 degrees
Three colorless solutions in test tubes, with no labels, are in a test tube rack on the laboratory bench. Lying beside the tests tubes are three labels : 0.10 M Na2CO3, 0.10 M HCL, and 0.10 M KOH. You are to place the labels on the test tubes using only the three solutions present. Here are your tests:
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm.
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Identify the labels for test tubes 1, 2, and 3
Answer:
Test tube 1 0.10 M HCL
Test tube 2 0.10 M KOH
Test tube 3 0.10 M Na2CO3
Explanation:
From the question we are told that
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm
Generally this warmth is as a result of a reaction between an acid and a base and the acid is 0.10 M HCL and the base is 0.10 M KOH , the heat generated is know as the heat of neutralization,
The reaction is
[tex]HCl_{(aq)} + KOH_{(aq)} \rightarrow KCl_{(aq)} + H_2O_{(l)} + \Delta H[/tex]
We are also told from the reaction that
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Generally carbon dioxide gas is produced is as a result of a reaction between the acid HCl and Na2CO3.
The reaction is
[tex]2HCl -{(aq)} + Na_2 CO_3_{(aq)} \rightarrow 2 NaCl _{(aq)} + CO_2_{(g)} + H_2O_{(l)}[/tex]
Hence from this explanation above we see that the solution in test tube 1 is 0.10 M HCL while solution in test tube 2 is 0.10 M KOH and then solution in test tube three is 0.10 M Na2CO3
4. What reagent would you predict to be in excess for reacting 7.50 mL of a 0.10M BaCl2 solution with 7.50 mL of 0.10M KIO3 solution
Answer : [tex]BaCl_2[/tex] reagent predict to be in excess.
Explanation : Given,
Concentration of [tex]BaCl_2[/tex] = 0.10 M
Volume of [tex]BaCl_2[/tex] = 7.50 mL = 0.0075 L (1 L = 1000 mL)
Concentration of [tex]KIO_3[/tex] = 0.10 M
Volume of [tex]KIO_3[/tex] = 7.50 mL = 0.0075 L
First we have to calculate the moles of [tex]BaCl_2[/tex] and [tex]KIO_3[/tex].
[tex]\text{Moles of }BaCl_2=\text{Concentration of }BaCl_2\times \text{Volume of }BaCl_2[/tex]
[tex]\text{Moles of }BaCl_2=0.10M\times 0.0075L=0.00075mol[/tex]
and,
[tex]\text{Moles of }KIO_3=\text{Concentration of }KIO_3\times \text{Volume of }KIO_3[/tex]
[tex]\text{Moles of }KIO_3=0.10M\times 0.0075L=0.00075mol[/tex]
Now we have to calculate the excess and limiting reagent.
The balanced equilibrium reaction will be:
[tex]BaCl_2+2KIO_3\rightleftharpoons Ba(IO_3)_2+2KCl [/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]KIO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]
So, 0.00075 moles of [tex]KIO_3[/tex] react with [tex]\frac{0.00075}{2}=0.000375[/tex] moles of [tex]BaCl_2[/tex]
From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KIO_3[/tex] is a limiting reagent and it limits the formation of product.
Hence, [tex]BaCl_2[/tex] reagent predict to be in excess.
What does reflection mean?
Answer:
the throwing back by a body or surface of light, heat, or sound without absorbing it.
"the reflection of light"
Explanation:
i hope this helps
A bicycle rider is applying a force of 20 N while heading south against a wind blowing from the south with a force of 5 N
Answer:
the bike will go down 5 N
what is the point of doing an experiment. Help thank you
Answer:
to find out how somethings work
Explanation:
Answer:
Doing experiments is good because when you try these possibilities you can learn about something that you've tried or see why this experiment didn't really work. It also helps you understand that not everything you try will work. that's why we experiment.
Explanation:
Hope this helps:)
A piece of metal has a mass of 0.650 kilograms, has a width of 0.136 meters, and has a length of 0.0451 meters.Part A: If the metal’s volume is 291 cm3, what is the height of the metal in centimeters? (The width & length values given above are in a different unit!)
Part B: What is the density of this piece of metal?
Answer:
height = 4.74 cm
density = 2.23 g/ cm³
Explanation:
Mass of metal = 0.650 kg (650 g)
Width = 0.136 m
Length = 0.0451 m
Volume of metal = 291 cm³
Height in cm = ?
density of metal =?
Solution:
Width = 0.136 m (0.136 m×100 cm/1m = 13.6 cm)
Length = 0.0451 m (0.0451 m×100 cm/1m = 4.51 cm)
First of all we will calculate the height:
Volume = height× width× length
291 cm³ = h × 13.6 cm × 4.51 cm
291 cm³ = h × 61.34 cm²
h = 291 cm³ / 61.34 cm²
h = 4.74 cm
Density:
d = m/v
d = 650 g/291 cm³
d = 2.23 g/ cm³
2) (3 pts) Convert 85 oF to oC.
Answer:
The answer will be 29.4˚C.
Explanation:
Using the formula 5(˚F-32)/9, plug in the numbers and you'll get 29.4˚C.
Which element will gain three electrons to form an anion?
A. aluminum
B. chromium
C. iron
D. nitrogen
Answer:
D represents the element nitrogen which will gain three valence electrons forming a 3 ion.
Answer:
answer is
Explanation:
D
When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq) Ag+ (aq) + I- (aq) → AgI (s) AgNO3 (aq) + KI (aq) → AgI (aq) + KNO3 (s) Ag+ (aq) + NO3 - (aq) → AgNO3 (aq) Ag+ (aq) + NO3 - (aq) → AgNO3 (s)
Answer:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Explanation:
We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.
Aqueous AgNO₃ and KI will dissociate in solution as follow:
AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)
KI (aq) —> K⁺(aq) + I¯(aq)
Aqueous AgNO₃ and KI will react as follow:
AgNO₃ (aq) + KI (aq) —>
Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)
Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.
Thus, the net ionic equation is:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
The net ionic equation of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex] to form [tex]AgI[/tex] precipitates is: B. [tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]
A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.
This ultimately implies that, both the charge on each atom and sum of the masses of the chemical compounds or elements in a chemical equation are properly balanced.
An ion can be defined as an atom or molecules (group of atoms) that has lost or gained one or more of its valence electrons, thereby, making it have a net positive or negative electrical charge.
First of all, we would write the dissociation equation for aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:
For [tex]AgNO_3[/tex]:
[tex]AgNO_3_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]
For [tex]KI[/tex]:
[tex]KI_{(aq)}[/tex] -----> [tex]K^{+}_{(aq)} + I^{-}_{(aq)}[/tex]
Next, we would write a chemical equation for the reaction of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:
[tex]AgNO_3_{(aq)} + KI_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex] [tex]+ \;K^{+}_{(aq)} + I^{-}_{(aq)}[/tex] ----->[tex]AgI_{(s)} + K^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]
Note: Spectator ions refers to the ions that exist as a reactant and a product in a chemical equation because they are unchanged by the chemical reaction.
In this chemical reaction, the spectator ions are:
[tex]K^+[/tex][tex]NO_{3}^{-}[/tex]Finally, in order to obtain the net ionic equation, we would cancel out the two (2) spectator ions:
[tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]
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