Answer:
quantity of heat=mc*theta
=25*0.930(76-25)
=25*0.930*51
=1185.75J
=11.9kJ
Answer:
1.19kJ
Explanation:
The heat can be calculated using the specific heat capacity, mass, and the change in temperature.
q=mxcxΔT=(25.0g)×(0.930Jg⋅∘C)×(76.0−25.0∘C)=1186J
Finally, the heat can be converted from J into kJ using the metric conversion factor.
1186J×1kJ1000J=1.19kJ
Phosphorus-32 has a half-life of 14.0 days Starting with 4.00 g of 2P how many grams will remain after 112 days?
Answer:
You'll be left with 0.25 g of phosphorus-32. The nuclear half-life of a radioactive isotope expresses the time needed for a sample of that isotope to reach half of its initial value. In your case, regardless with how much phosphorus-32 you start with, you'll be left with half of that initial mass after 14.0 days.
Explanation:
Ethyl bromide reacts with sodium methoxide to form ?
Answer:
The reactants should be ethyl bromide and sodium methoxide.
The products of the reactions are diethylether and sodium bromide.
Explanation:
When ethyl bromide, it reacts with any sodium ethoxide, it gives diethylether and also sodium bromide.
As in the sodium ethoxide, the negatively charged oxygen atom is very much reactive and it attacks on the bromine containing a carbon of the ethyl bromide and it replaces it forming methyl ethoxide. This type of reaction is sometimes known as SN2.
2. (2 points) Calculate the Effective Nuclear Charge for each atom. Show all of your work for full credit. Se: Br: (1 point) Which atom is larger? (This is your claim.)
Answer:
Bromine.
Explanation:
There is more nuclear charge on bromine because in bromine, there are 35 number of protons and 46 number of neutrons while in selenium, there are 34 number of protons and 45 number of neutrons. The atomic size of selenium is larger because they have lower nuclear charge as compared to bromine and we know that when we go to the left of periodic table atomic size increases which attract the electrons towards itself so the shell comes close to the nucleus and decrease occurs in atomic radius.
In order to derive a simplified version of the Nernst equation by assuming standard temperature we can substitute which of the following into the general equation?
a. Faraday's constant
b. Standard pressure (1 atm)
c. 298 K
d. 25 °C
Answer:
c. 298 K
Explanation:
Nernst equation is an equation used in electrochemistry that relates the reduction potential of a reaction with the standard potential, temperature and concentrations of the reactants in that are been reducted and oxidized. The formula is:
E = E° - RT / nF ln [Red] / [Ox]
Where R is gas constant (8.314J/molK), T is absolute temperature (In Kelvin), n are moles of electrons and F is faraday constant (K/Volt*mol)
In electrochemistry, standard temperature is taken as 298K. That means by assuming standard temperature we can substitute T as:
c. 298 KArrange these reactions according to increasing ΔS.
a. H 2O(g) → H 2O(l)
b. 2NO(g) → N 2(g) + O 2(g)
C. MgCO 3(s) → MgO(s) + CO2(g)
Explanation:
ΔS denotes change in entropy. Entropy is the degree of disorderliness of a system. An increasing ΔS would mean that the entropy of the products is greater than the entropy of the reactants.
Generally the trend of entropy with state of matter is given as;
Gas > Liquid > Solid (Increasing Entropy)
a. H 2O(g) → H 2O(l)
In this reaction, there is a decreasing ΔS, since we are moving from gas to liquid. This would be the last.
b. 2NO(g) → N 2(g) + O 2(g)
There is an increase in gaseous products. This is definitely an increasing ΔS reaction. This would be the first on the list.
C. MgCO 3(s) → MgO(s) + CO2(g)
The products formed are in solid and gaseous state. There is increasing ΔS but not up to reaction 2.
The order is given as;
b. 2NO(g) → N 2(g) + O 2(g)
C. MgCO 3(s) → MgO(s) + CO2(g)
a. H 2O(g) → H 2O(l)
Glade air freshener gel “disappearing” is an example of
Answer:
Vaporization
Explanation:
Vaporization is the change of a specie to the gaseous state. To 'disappear' in this case simply means to change to the gaseous state.
Substances with high vapour pressure tend to be easily converted to vapour phase. Hence if Glade air freshener gel 'disappears' easily, then it has a high vapour pressure and is easily converted to vapour (gas).
Organic molecules are defined as chemical compounds that contain ______ in distinct ratios and structures.
Answer:
The answer is: carbon
Explanation:
Organic molecules contain the chemical element carbon (C) in its structure. In this type of molecules, carbon is usually bonded to hydrogen (H), oxygen (O) and, with less frecuency, nitrogen (N). Therefore, in these molecules, carbon forms simple, double and triple bonds with itself. Examples of organic molecules that are very important in biology are carbohydrates, lipids, proteins and nucleic acids.
Cell notation will list each half-reaction:
A.) starting with the reaction at the cathode
B.) with coefficients to balance the species
C.) on each side of the salt bridge, which is represented by a double vertical line
D.) none of the above
Answer:
on each side of the salt bridge, which is represented by a double vertical line
Explanation:
While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.
The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.
Of the following, which is the dominant (strongest) type of intermolecular force present in RbCl(s)?
A. lon-dipole
B. lonic
C. Dispersion
D. Dipole-dipole
E. Hydrogen bonding
Answer:
B. Ionic
Explanation:
RbCl is ionic compound formed from the combination of rubidium (a strongly electropositive element) and chlorine (a strongly electronegative element).
The transfer of an electron from rubidium to chlorine leads to the formation of two oppositely charged ions Rb⁺ and Cl⁻.
In a solid crystal of RbCl, the major intermolecular forces of attraction are electrostatic forces of attraction between oppositely charged ions in the molecules of the crystalline solid.
Ion-dipole attraction only occur when the solid is dissolved in a polar molecule like water. Dispersion forces as well as dipole-dipole forces, though present are weak forces and contribute very little to intermolecular attraction. Hydrogen bonding is not present as there are no hydrogen atoms in solid RbCl.
The dominant (strongest) type of intermolecular force present in RbCl(s) is ionic force (Option B).
An ion-dipole attraction refers to a type of attraction between ions and neutral dipole molecules. Ionic forces are commonly observed in different types of chemical solutions. An ion-dipole force consists of a positive/negative ion that exerts an attractive force to the negative/positive end present in a polar molecule.In conclusion, the dominant (strongest) type of intermolecular force present in RbCl(s) is ionic force (Option B).
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Sort the following gases by whether they are transparent to incoming short-wavelength UV radiation or whether they absorb longer wavelength terrestrial re-radiation.
O2
N2
CH4
H2O
CO2
Ar
Categories
Absorb No Radiation Absorb Re-Radiation
Answer:
Short wavelength UV radiation ----Oxygen (O2 ), nitrogen (N2) , methane (CH4) , water (H2O) and carbondioxide (CO2).
longer wavelength radiation---- methane (CH4) , water (H2O) and carbondioxide (CO2)
Explanation:
Oxygen (O2 ), nitrogen (N2) , methane (CH4) , water (H2O) and carbondioxide (CO2) are the gases which are transparent for incoming short-wavelength UV radiation while methane (CH4) , water (H2O) and carbondioxide (CO2) are the greenhouse gases which absorb longer wavelength. When the solar radiation comes to the earth, it has short wavelength but after reflection it converts into longer wavelength which can be prevent by greenhouse gases and increases the temperature of atmosphere.
CH4, H2O and CO2 gases absorb longer wavelength terrestrial re-radiation.
UV radiation are short wavelength radiation that reach the earth from outer space. However, certain gases that compose the atmosphere are able to absorb longer wavelength terrestrial re-radiation.
The following gases absorb longer wavelength terrestrial re-radiation;
CH4H2OCO2The following gases are transparent to incoming short-wavelength UV radiation;
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Relative to the energy stored in chemical bonds, nuclear binding energies are:_________.
a. much greater
b. much smaller
c. approximately equal
d. depends on the substance
Answer:
much greater
Explanation:
Binding energy of a nucleus is the energy evolved in the formation of a nucleus from free protons and neutrons. It is also defined as the energy that binds nucleons together in the nucleus of an atom.
The nuclear binding energy is far greater than the chemical energy stored in chemical bonds. Nuclear energy is the greatest source of energy which could be exploited to achieve various ends.
Which of the following is not a technique that could be used to separate a mixture into its components?
a) stirring.
b) decanting.
c) filtration.
d) distillation.
e) none of the above.
Answer:
Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as evaporation, distillation, filtration and chromatography.
Explanation:
You are very much welcom
Answer:
stirring. because it will mixup together
Consider the reaction when aqueous solutions of potassium hydroxide and ammonium nitrate are combined. The net ionic equation for this reaction is:
Answer: [tex]OH^-(aq)+NH_4^+(aq)\rightarrow NH_3(g)+H_2O(l)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex]KOH(aq)+NH_4NO_3(aq)\rightarrow KNO_3(aq)+NH_3(g)+H_2O(l)[/tex]
The equation can be written in terms of ions as:
[tex]K^+(aq)+OH^-(aq)+NH_4^+(aq)+NO_3^-(aq)\rightarrow K^+(aq)+NO_3^-(aq)+NH_3(g)+H_2O(l)[/tex]
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The ions which are present on both the sides of the equation are potassium and nitrate ions and hence are not involved in net ionic equation.
Hence, the net ionic equation is :
[tex]OH^-(aq)+NH_4^+(aq)\rightarrow NH_3(g)+H_2O(l)[/tex]
Which process is an example of a chemical change?
Answer:
burning of paper
melting of lead
Answer:
d- activating a glow stick
Explanation:
Glow sticks produce light through a chemical reaction.
The molar solubility of a slightly soluble ionic compound M2X3 is 2.8 x 10-6 M. Determine the value of Ksp.
Answer:
1.9 × 10⁻²⁶
Explanation:
Step 1: Write the solution reaction for M₂X₃
M₂X₃(s) ⇄ 2 M³⁺(aq) + 3 X²⁻(aq)
Step 2: Make an ICE chart
We can relate the molar solubility (S) with the solubility product constant (Ksp) using an ICE chart.
M₂X₃(s) ⇄ 2 M³⁺(aq) + 3 X²⁻(aq)
I 0 0
C +2S +3S
E 2S 3S
The solubility product constant is:
Ksp = [M³⁺]² × [X²⁻]³ = (2S)² × (3S)³ = 108 S⁵ = 108 (2.8 × 10⁻⁶)⁵ = 1.9 × 10⁻²⁶
The value of Ksp when there is the slightly soluble ionic compound so it should be 1.9 × 10⁻²⁶.
Calculation of the value of ksp:Since the solution reaction for M₂X₃ should be
M₂X₃(s) ⇄ 2 M³⁺(aq) + 3 X²⁻(aq)
Now make an ICE chart
So it can be like
M₂X₃(s) ⇄ 2 M³⁺(aq) + 3 X²⁻(aq)
I 0 0
C +2S +3S
E 2S 3S
Now The solubility product constant is:
Ksp = [M³⁺]² × [X²⁻]³
= (2S)² × (3S)³ = 108 S⁵
= 108 (2.8 × 10⁻⁶)⁵
= 1.9 × 10⁻²⁶
hence, The value of Ksp when there is the slightly soluble ionic compound so it should be 1.9 × 10⁻²⁶.
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What is the atom inventory for the following equation after it is properly balanced? ____K2S + ____CoCl2 Imported Asset ____KCl + ____CoS
Reactants: K = 2, S = 1, Co = 1, Cl = 2; Products: K = 2, S = 1, Co = 1, Cl = 2 Reactants: K = 2, S = 2, Co = 1, Cl = 2; Products: K = 2, S = 1, Co = 1, Cl = 1 Reactants: K = 2, S = 2, Co = 1, Cl = 2; Products: K = 2, S = 2, Co = 1, Cl = 1 Reactants: K = 2, S = 1, Co = 2, Cl = 2; Products: K = 2, S = 1, Co = 1, Cl = 2
Answer:
Reactants: K = 2, S = 1, Co = 1, Cl = 2; Products: K = 2, S = 1, Co = 1, Cl = 2
Explanation:
The reaction is a double replacement reaction so the anions (Cl⁻ and S²⁻) switch places.
1 K₂S + 1 CoCl₂ ⇒ 2 KCl + 1 CoS (balanced chemical equation)
On the reactants and products side, K = 2, S = 1, Co = 1, and Cl = 2.
Hope that helps.
2. Who was the first Avenger to crumble to dust following Thanos' snap?
A. Captain America
B. Black Panther
C. Bucky Barnes
D. The Hulk
Answer:
C. bucky barnes
Explanation:
i dont know how this relates to chemistry
Select the conjugate acid-base pair(s). a) HI, I b) HCHO2, SO4^2- c) CO3^2-, HCI d) PO4^3-, HPO4^2-
Answer:
PO4^3-, HPO4^2-
Explanation:
If an acid gives out a proton, the acid then changes to its corresponding base. Similarly, if a base takes in a proton, it changes to its corresponding acid. If a pair of acid and base differ only by the presence or absence of a proton, then they are referred to as a conjugate acid-base pair.
Let us look at this;
HPO4^2-(aq) ----> H^+(aq) + PO4^3-(aq)
HPO4^2- and PO4^3- differ only in the presence or absence of a proton (H^+) hence they constitute a conjugate acid-base pair.
Draw the Lewis structure for the polyatomic trisulfide anion. Be sure to include all resonance structures that satisfy the octet rule.
Answer:
Explanation:
The objective here is to draw the Lewis structure for the polyatomic trisulfide anion and to be sure all resonance structures that satisfy the octet rule are included.
The Lewis structure for Polyatomic trisulfide anion
The first step is to the layout the skeleton of the Polyatomic trisulfide anion
S S S
However, the next step is to make sure we fill in the bonding pairs of electrons on the central atom.
Then , we move over to filling the lone pairs electrons before we finally have the Lewis structure for Polyatomic trisulfide anion as shown in the image below.
Convert
propene n to propane
⬇️
Explanation:
First convert propane to chloropropane by halogenation reaction in presence of sunlight
CH3-CH2-CH3 + Cl2 → CH3-CH2-CH2Cl + HCl
Chloropropane can be converted into propene by using alcoholic KOH
CH3-CH2-CH2Cl + alc. KOH → CH3-CH=CH2
To convert from Propene to Propan-2-ol, the addition of H2SO4 takes place in accordance with Markovnikov ‘s rule i.e.
Hydrogen can be prepared on a small scale by the hydrolysis of metal hydrides:
CaH2 + 2H20 — Ca(OH)2 + 2H2
Assign an oxidation number to each atom and verify that the total number of electrons lost is equal to the total number gained.
Answer:
Ca²⁺ + 2H⁻ + 2(2H⁺ + O²⁻) ----> Ca²⁺ + 2O²⁻ + 2H⁺ + 4H⁰
(2 electrons lost) ----> (2 electrons gained)
Explanation:
The equation of the reaction is as follows: CaH₂ + 2H₂0 —> Ca(OH)₂ + 2H₂
The oxidation numbers of each of the atoms in the molecules are as follows:
In CaH₂, calcium has an oxidation number of +2, while hydrogen has an oxidation number of -1 ; Ca²⁺, 2H⁻
In 2H₂0, hydrogen has an oxidation number of +1 while oxygen has an oxidation number of -2 ; H⁺, O²⁻
In Ca(OH)₂, calcium has an oxidation number of +2 while oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1 ; Ca²⁺, O²⁻, 2H⁺
In 2H₂, each hydrogen atom has an oxidation number of 0 ; 4 H⁰
Ca²⁺ + 2H⁻ + 2(2H⁺ + O²⁻) ----> Ca²⁺ + 2O²⁻ + 2H⁺ + 4H⁰
On the left hand side, the hydride ion gives up two electrons to become neutral hydrogen atoms.
On the right hand, two hydrogen atoms accept two electrons to form neutral atoms.
These neutral hydrogen atoms pair up to form two hydrogen gas molecules.
A
Identify the exact numbers.
2.54 cm (the number of centimeters in exactly 1 in)
4 choices (the number of choices in this list)
56.00 s (a time measured by a stopwatch)
3 lbs (the weight of an object measured on a scale)
Answer:
the answer is (2.54cm in an inch, and 4 number of choices in this question)
Explanation:
An exact number is not rounded, and cannot be changed such as placing a different item on the scale it would change.
Two function of mass spectrometry
Your question has been heard loud and clear.
Specific applications of mass spectrometry include drug testing and discovery, food contamination detection, pesticide residue analysis, isotope ratio determination, protein identification, and carbon dating.
Thank you
The primary function of mass spectrometry is as a tool for chemical analyses based on detection and quantification of ions according to their mass-to-charge ratio. However, mass spectrometry also shows promise for material synthesis.
tbh google
✌️
Solid sodium reacts with liquid water to form hydrogen gas according to the equation 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) What is the pressure of hydrogen gas in the 20.0 L headspace of a reactor vessel when 3.34 kg sodium is reacted with excess water at 50.0°C?
Answer:
96.2 atm
Explanation:
Step 1: Write the balanced equation
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
Step 2: Calculate the moles corresponding to 3.34 kg of sodium
The molar mass of Na is 22.99 g/mol.
[tex]3.34 \times 10^{3} g \times \frac{1mol}{22.99g} =145mol[/tex]
Step 3: Calculate the moles of hydrogen formed from 145 moles of sodium
The molar ratio of Na to H₂ is 2:1. The moles of H₂ formed are 1/2 × 145 mol = 72.5 mol
Step 4: Convert the temperature to the Kelvin scale
K = °C + 273.15 = 50.0°C + 273.15 = 323.2 K
Step 5: Calculate the pressure exerted by the hydrogen gas
We will use the ideal gas equation.
[tex]P \times V = n \times R \times T\\P = \frac{n \times R \times T}{V} = \frac{72.5mol \times 0.0821atm.L/mol.K \times 323.2K}{20.0L} = 96.2 atm[/tex]
Atomic radius generally increases as we move
A) down a group and from right to left across a period
B) up a group and from left to right across a period
C) down a group and from left to right across a period
D) up a group and from right to left across a period
E) down a group; the period position has no effect
Answer:
C
Explanation:
Atomic radius generally increases as we move down a group and from left to right across a period
how many molecule of carbon dioxide are needed to react with excess iron oxide to produce 11.6 g of iron
Answer:
0.16 moles of Carbon
Explanation:
The balanced reaction equation:
[tex]2Fe_{2}O_{3}[/tex] + [tex]3C[/tex] → [tex]4Fe[/tex] + [tex]3CO_{2}[/tex]↑
The mole ratio of Carbon to Iron is 3 : 4 (since Fe2O3 is in excess)
i.e 3 moles of C produces 4 moles of Fe.
If 1 mole of Fe - 55.8g of Fe
? moles - 11.6g of Fe
= [tex]\frac{11.6}{55.8}[/tex] = 0.208 moles
But 3 moles of C - 4 moles of Fe
? moles of C - 0.208 moles of Fe
= [tex]\frac{3 *0.208}{4}[/tex] = 0.16 moles of carbon.
I hope this explanation was clear and useful.
A force of 26.4lb is applied to an area of 1.71 in2. What is the pressure?
A) 15.4 lb/in2
B) 0.0648 lb/in2
C) 45.1 lb/in2
D) 22.8 lb/in2
Answer:
Pressure, P = 15.4 lb/in²
Explanation:
Given that,
Force applied, F = 26.4 lb
Area of cross section, A = 1.71 in²
To find,
Pressure.
Solution,
Pressure is equal to force per unit area of cross section.
[tex]P=\dfrac{F}{A}\\\\P=\dfrac{26.4\ lb}{1.71\ in^2}\\\\P=15.43\ lb/in^2\\\\P=15.43\ Pa[/tex]
Hence, the correct option is (a) "15.4 lb/in²".
Vitamin C ascorbic acid (C6F1806), is a reducing agent. One way to determine the ascorbic acid content of a sample is to mix the acid with an excess of iodine
C6H8O6 + I2 + H2O------- C6H6O6 + 2H3O + 2I-
and then titrate the iodine that did not react with the ascorbic acid with sodium thiosulfate. The balanced, net ionic equation for the reaction occurring in this titration is
I2+ 2 S2O32-(a9) —) 2 I-(a9) + S4O62-(a9)
Suppose 50.00 mL of 0.0520 M 12 was added to the sample containing ascorbic acid. After the ascorbic acid/I2 reaction was complete, the 12 not used in this reaction required 20.30 mL of 0.196 M Na2S2O3 for titration to the equivalence point. Calculate the mass of ascorbic acid in the unknown sample.
Answer:
The mass of ascorbic acid in the sample is 0.1076g
Explanation:
In other words to determine ascorbic acid in a sample:
You are adding an amount of I₂. One part is reacting with ascorbic acid and the other with sodium thiosulphate. First, we need to calculate how much thiosulphate reacts to know the iodine that reacts with thiosulphate.
I₂ that reacts with S₂O₃²⁻:
Moles S₂O₃⁻: 0.02030L × (0.196mol / L) = 3.979x10⁻³ moles S₂O₃²⁻.
As 2 moles of S₂O₃²⁻ reacts per moles of I₂, moles of I₂ are:
3.979x10⁻³ moles S₂O₃²⁻ ₓ (1 mole I₂ / 2 moles S₂O₃²⁻) = 1.989x10⁻³ moles I₂
Now, initial amount of I₂ added to the sample is:
0.0500L ₓ (0.0520mol I₂ / L) = 2.600x10⁻³ moles I₂
That means moles of I₂ that react with ascorbic acid = Moles of ascorbic acid are:
2.600x10⁻³ moles I₂ - 1.989x10⁻³ moles I₂ =
6.11x10⁻⁴ moles I₂ = Moles C₆H₈O₆.
To convert moles of substance to mass we need to use molar mass of the substance (176.12g/mol for ascorbic acid):
6.11x10⁻⁴ moles C₆H₈O₆ ₓ (176.12g / mol) =
The mass of ascorbic acid in the sample is 0.1076gAt elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen:_________.
2N2O5(g)→4NO2(g)+O2(g)
When the rate of formation of O2 is 8.1 x 10^−4 M/s, the rate of decomposition of N2O5 is _____ M/s.
a. 3.2 x 10^−3
b. 8.1 x 10^−4
c. 1.6 x 10^−3
d. 2.0 x 10^−4
e. 4.1 x 10^−4
Answer:
e. 4.1 x 10^−4
Explanation:
For the reaction;
2N2O5(g) → 4NO2(g) + O2(g)
The rate of formation is given as;
(1 / 4) Δ [O2] / Δt = (1 / 2 )Δ [N2O5] / Δt
Δ [O2] / Δt = 8.1 x 10^−4 M/s
Inserting into he equation, we have;
(1/4) (8.1 x 10^−4 ) = (1/2) (Δ [N2O5] / Δt)
2.025 x 10^−4 = (1/2) (Δ [N2O5] / Δt)
Δ [N2O5] / Δt = 2 * 2.025 x 10^−4
Δ [N2O5] / Δt = 4.1 x 10^−4
Correct option is option E.
Students set up a controlled experiment. They put five
crickets in each of three identical containers. They set one
container to 15°C, one to 20°C, and one to 25°C. Then they
count the number of cricket chirps in each container
during 1 hour
What are two constants in this experiment?
A. The number of crickets in each container
B. The number of cricket chirps
O C. The kind of containers used
D. The temperature of each container
The experimental design includes the controlled, independent, and dependent variables. The number of crickets and kinds of containers is two constant. Thus, options A and C are correct.
What are control variables?Control variables are the experimental element that remains fixed and unchanged in an experimental setup. They are constant and not are directly related to the interest of the investigation topic but are known to influence the results and observations.
The control variables can influence the dependent variable of the experiment as the number of crickets in each container and the type of container can indirectly alter the outcome and the conclusion of the experiment. The number of cricket will influence the effect of the temperature and the type of container can alter the survival rate.
Therefore, in options A and C. the number of cricket and container are two constants.
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