The following data were obtained using the same procedure as in your experiment. Using these data, calculate the percent malachite in the sample.
FW information (in g/mol): malachite = 221.11; CO2 = 44.01; H2O = 18.02
Mass of crucible: 23.8839 g Mass of crucible + sample: 26.0496 g
Mass of crucible + decomposed sample: 25.7013 g
One molecule of malachite decomposes to make:_______.
___________molecule(s) of copper oxide
___________molecule(s) of carbon dioxide
____________molecule(s) of water
a. 1, 2,2
b. 2, 1, 1
c. 1, 1, 1
d. 1,0.5, 0.5

Answers

Answer 1

Answer:

c

Explanation:


Related Questions

Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?

Answers

cyclohexane will distill off first as it will have lower boiling point compared to ortho xylene which has higher molecular mass

what is valency of an atom?​

Answers

The number of replaceable electrons in an atom is called its valency.

Examples

Monovalent - HydrogenDivalent - Oxygen

Valency = 8 - Number of electron in last shell [When number of electrons in last shell > 4]

Valency = Number of electron in last shell [When number of electrons in last shell < 4]

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Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

the nutrition label on rice lists the amounts of protein, carbohydrates and fats in one serving. these substances are important for human nutrition

Answers

Answer:

Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.

Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.

Explanation:

state function and non state function ​

Answers

Answer:

State functions represent quantities or properties of a thermodynamic system, while non-state functions represent a process during which the state functions change. For example, the state function PV is proportional to the internal energy of an ideal gas, but the work W is the amount of energy transferred as the system performs work.

Explanation:

Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email

Answers

Answer:

d

Explanation:

since it is much convenient since the email will not get lost and it's contents will not be forgotten

g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular equation, what are the products

Answers

Answer:

Where the products are H2O and Ba(NO3)2

Explanation:

A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called neutralization reaction.

The balanced reaction is:

Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2

Where the products are H2O and Ba(NO3)2

An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize

Answers

Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

Moles HCl added:

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

Moles NaOH to titrate the excess:

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

Moles of acid that were neutralized:

0.0105 moles - 0.0010 moles =

0.0095 moles of acid were neutralized by the antiacid

Starting from (R)-3-methylhex-1-yne as the substrate at the center of your page, draw a reaction map showing the regiochemical and stereochemical outcome or outcomes for each of the following series of reagents. Name each of your products, including stereochemical designations for any chirality centers that are generated.

a. HgSO4, H2SO4, H2O
b. 1. 9-BBN; 2. H2O2, NaOH
c. Br2, CCl4
d. HBr

Answers

Solution :

A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.

Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.

In the context, the products including the stereochemical designations for any chirality centers starting from the  (R)-3-methylhex-1-yne as the substrate are attached below.  

cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?

Answers

1.37% of cesium–131 will remain after 60 days

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Percentage remaining after 60 days =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Number of half-lives (n) =?

n = t / t½

n = 60 / 9.7

Finally, we shall determine the percentage remaining. This can be obtained as follow:

Let the original amount be N₀

Let the amount remaining be N

Number of half-lives (n) = 60 / 9.7

N = N₀ / 2ⁿ

Divide both side by N₀

N/N₀ = 1/2ⁿ

N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾

N/N₀ = 0.0137

Multiply by 100 to express in percentage

N/N₀ = 0.0137 × 100

N/N₀ = 1.37%

Therefore, the percentage remaining after 60 days is 1.37%

NOTE; N/N₀ is the fraction remaining.

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Determine the value of the equilibrium constant for the following reaction, where the following amounts of each species are present at equilibrium in a 5.00 L container: 1.34 mol HCl, 4.30 mol O2, 30 g H2O, and 2.42 mol Cl2.
4 HCl(g) O2(g) ----> 2 H2O(l) 2 Cl2(g)

Answers

Explanation:

here's the answer to your question about

Which redox reaction would most likely occur if silver and copper metal were added to a solution that contained silver and copper ions?
A. Cu + Agt Cu2+ + 2Ag
B. Cu2+ + 2Ag* → Cu + 2Ag
C. Cu2+ + 2Ag → Cu + 2Ag+
D. Cu + 2Ag Cu²+ + 2Ag+​


give the wrong answer and I'm reporting ​

Answers

Answer:

B

Explanation:

b/c copper is readuction agent

The most likely redox reaction that would occur if silver and copper metal were added to a solution that contained silver and copper ions is   [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]. The correct answer is option C.

Redox reaction is a reaction in which reduction and oxidation takes place simultaneously.

In this reaction:

[tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]

Copper metal has a higher reduction potential than silver metal, which means that it will be oxidized to [tex]\rm Cu^{2+}[/tex] ions before silver metal is oxidized to  [tex]\rm Ag^+[/tex] ions.

The [tex]\rm Cu^{2+}[/tex] ions in the solution will then react with the silver metal to form [tex]\rm Ag^+[/tex] ions and Copper metal. This reaction is an example of a displacement reaction, where a more reactive metal removes a less reactive metal from its compound.

Therefore, option C. [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex] is the correct answer.

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Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to one compartment of the microwell plate. She then makes a solution of MCln by adding KCl to M(NO3)n. She adds 7.903 mL of the resulting mixture to a second compartment of the microwell plate. Sally knows n = +2. She has already calculated [Mn+] in the prepared MCln solution using the Nernst equation. [Mn+] = 8.279 M

Required:
How many moles of [Cl-] must be dissolved in that compartment?

Answers

Answer:

0.1309 mol

Explanation:

From the given information:

The metal ion, two ions of [tex]M^{+}[/tex] reacted with Cl⁻ to form [tex]MCl_n[/tex] i.e. the compound formed is [tex]MCl_2[/tex].

The concentration of the metal ion formed [tex][M^+][/tex] = 8.279 M

The concentration of the chlorine ion formed [tex][Cl^-][/tex] = 2 × 8.279 M

= 16.558 M

We know that:

[tex]\mathsf{Molarity = \dfrac{no \ of \ moles }{volume (mL)}}[/tex]

The number of moles of [tex][Cl^-][/tex] = [tex]16.558 \ mol.L^{-1} \times 7.903 \ mL \times \dfrac{1 \ L}{1000 \ mL}[/tex]

= 0.1309 mol

PLEASE HELP ASAP
A total of 132.33g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed.
• How many grams of CO2 and H2O will be produced? (2 points)
















b. If the furnace is not properly adjusted, the products of combustion can include other gases, such as CO and unburned hydrocarbons. If only 269.34 g of CO2 were formed in the above reaction, what would the percent yield be? (2 points)

Answers

A combustion reaction takes place in the presence of O2 (g). The
products of a complete combustion reaction are CO2 (g) and H2O
(g).

2.5
Nakula investigated the effect of heat on amylase. Amylase is an enzyme
that makes starch molecules break down into sugar molecules.
P
• Nakula put some amylase solution into two boiling tubes,
P and Q
• He boiled the solution in tube P. He did not heat tube Q.
• He waited until the solution in tube P had cooled down to
room temperature.
• He added equal volumes of starch solution to tube P and
boiled amylase amylase
• After 10 minutes, he tested both tubes for sugar.
and starch and starch
• Nakula found that there was sugar in tube Q. but not in tube P.



Answers

The structure of amylase deteriorates due to high temperature of the solution.

This experiment shows that the structure of amylase deteriorates due to high temperature which prevents this amylase from performing its function properly.

At high temperatures the amylase will break starch down very slowly or not at all due to denaturation of the enzyme's active site due to which it can't perform its function properly so we can conclude that high temperature denatures amylase enzyme.

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Nakula's conclusion was- "My results show that boiling destroys amylase"

Amylase is an enzyme that breaks down starch molecules into sugar molecules. He boiled the solution in tube P, and when he checked tube P for sugar, there wasn't any. He didn't boil the solution in tube Q and he found sugar in it.

Amylase had broken down starch molecules to sugar molecules in tube Q. Tube P's solution had been boiled, and this showed that when he boiled it, it destroyed the amylase, that is why the starch molecules hadn't been broken down into sugar molecules.

what type of bonding does Sodium Sulphate comes under?and explain in detail please​

Answers

Answer:

The bond between sodium sulfate is an ionic bond since it's a bond between a metal and non metals however the bond between sulfur and oxygen is a covalent bond since the two are non metals and the other reason that makes this an ionic bond is that there is both losing and gaining of electrons..

I hope this helps

Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.

Answers

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.

How we calculate the [OH⁻]?

We can calculate the concentration of hydroxide ions as follow:

[OH⁻][H⁺] = 10⁻¹⁴

Given chemical reaction with ICE table shown as below:

                H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq)

Initial:             0.115                      0                    0

Change:           -x                        +x                  +x

Equilibrium:  0.115-x                   +x                  +x

Given that, Ka = 4.3 x 10⁻⁷

Equilibrium constant for this reaction is written as:

Ka = [H⁺][HCO₃⁻]/[H₂CO₃]

4.3 x 10⁻⁷ = x² / 0.115

x = 2.32 x 10⁻⁴M = [H⁺]

Now we calculate the concentration of hydroxide ion as:

[OH⁻][H⁺] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M

Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.

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Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.

a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.

1. Increase
2. decrease
3. No effect

Answers

Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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The decrease in the water table due to overuse of water.

Answers

Answer:  Groundwater and surface water are connected. When groundwater is overused, the lakes, streams, and rivers connected to groundwater can also have their supply diminished. Land subsidence occurs when there is a loss of support below ground. This is most often caused by human activities, mainly from the overuse of groundwater, when the soil collapses, compacts, and drops.

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A filament for a light bulb needs to conduct electricity. Which of the elements listed below might be useful as a light bulb filament? Explain your thinking.

A. Tungsten, W
B. Sulfur, S
C. Bromine, Br

Answers

Answer:

A. tungsten

Explanation:

Tungsten is a material which high melting point ie. does not melt easily incase of high temperature

Answer:

option(A):Tungsten

Explanation:

tungsten has highest melting point.

Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...

Answers

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

Explanation:

The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubility

So, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

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Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution. ​

Answers

Answer:

Molarity is 0.075 M.

Explanation:

Moles:

[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]

RFM of potassium sulphate :

[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]

substitute:

[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]

In volume of 5.00 l:

[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]

g An aqueous solution of nitric acid is standardized by titration with a 0.137 M solution of calcium hydroxide. If 19.0 mL of base are required to neutralize 21.8 mL of the acid, what is the molarity of the nitric acid solution

Answers

Answer:

M of HNO₃ is 0.119M

Explanation:

A basic concept of titration is that in equivalence point:

mmoles of acid = mmoles of base

We have data from base and we only have data from volume of acid.

In a case our titration is a strong acid against a strong base.

We apply formula:

M of acid . Vol of acid = M of base . Vol of base

M of acid . 21.8 mL = 0.137M . 19 mL

M of acid = (0.137M . 19 mL) / 21.8 mL

M of acid = 0.119 M

When we neutralize all the titrant we reach the equivalence point.

At this point, pH = 7

2HNO₃  +  Ca(OH)₂ → Ca(NO₃)₂ +  2H₂O

Not following hazardous material safety policies and procedures can result in which of
the following? (Select all that apply.)
a. Serious illnesses
b. Injury
c. Death
d. HIPAA violation

Answers

Answer:

A, B, C and D

Explanation:

It can result for all of the choices mentioned.

Not following safety and procedure for handling hazardous material results in illness, death, injury, and HIPAA violation. Thus, all options are correct.

The hazardous material safety policy and measures are the standards set by HIPAA for the safety and precautionary measures that have been followed for reducing personal risk.

The hazardous materials have been chemicals, gases, flammables, and explosives. The inappropriate handling and not following the standard procedure results in illness, injury, death, and HIPAA violation. Thus, all the options are correct.

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For the following acids of varying concentrations, which are titrated with 0.50 M NaOH, rank the acids in order of least to most volume of base needed to completely neutralize the acid.

a. 0.2M H2C6H5O7
b. 0.2M H2C2O4

Answers

Answer:

0.2M H2C6H5O7 < 0.2M H2C2O4

Explanation:

A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.

More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.

Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.

H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.

H₂C₆H₅O₇ < H₂C₂O₄

The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)

Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.

Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.

Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.

Strong acids ionizes completely to produce hydrogen ions.

Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.

In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.

H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23

H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08

Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇

For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.

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At a constant temperature, a sample of gas occupies 1.5 L at a pressure of 2.8 ATM. What will be the pressure of this sample, in atmospheres, if the new volume is 0.92 L?

Answers

V1=1.5LV2=0.92LP1=2.8atmP2=?

Using boyles law

[tex]\boxed{\sf v\propto \dfrac{1}{p}}[/tex]

[tex]\\ \sf\longmapsto P_1V_1=P_2V_2[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{P_1V_1}{V_2}[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{2.8\times 1.5}{0.92}[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{4.2}{0.92}[/tex]

[tex]\\ \sf\longmapsto P_2=4.56atm[/tex]

[tex]\\ \sf\longmapsto P_2\approx 4.6atm[/tex]

Answer:

[tex]\boxed {\boxed {\sf 4.6 \ atm}}[/tex]

Explanation:

We are asked to find the new pressure given a change in volume. 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_1V_1= P_2V_2[/tex]

Initially, the gas occupies 1.5 liters at a pressure of 2.8 atmospheres.

[tex]1.5 \ L * 2.8 \ atm = P_2V_2[/tex]

The volume is changed to 0.92 liters, but the pressure is unknown.

[tex]1.5 \ L * 2.8 \ atm = P_2* 0.92 \ L[/tex]

We are solving for the final pressure, so we must isolate the variable P₂. It is being multiplied by 0.92 liters. The inverse operation of multiplication is division, so we divide both sides by 0.92 L.

[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L} = \frac{P_2* 0.92 \ L}{0.92 \ L}[/tex]

[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L}= P_2[/tex]

The units of liters cancel each other out.

[tex]\frac {1.5 * 2.8 \ atm}{0.92 }=P_2[/tex]

[tex]\frac {4.2}{0.92} \ atm= P_2[/tex]

[tex]4.565217391 \ atm = P_2[/tex]

The original measurements of pressure and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 5 up to a 6.

[tex]4.6 \ atm \approx P_2[/tex]

The pressure is approximately 4.6 atmospheres.

Compare the modern (electron cloud) model of the atom with Bohr’s atomic model. Which of these statements describe the two models correctly? Check all of the boxes that apply.

A. Bohr’s model was replaced only because of its age.

B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model the location of the electrons cannot be predicted.

C. The modern model explains all available data about atoms; Bohr’s model does not.

D. The modern model is more widely accepted because it was proposed by more well known scientists.

Answers

Answer:

B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model, the location of the electrons cannot be predicted.

AND

C. The modern model explains all available data about atoms; Bohr’s model does not.

Explanation:

The answers are right on Edge. :)

Answer:

b and c

Explanation:

my assignment was 100% 2022

Which phenomenon explained below is an example of deposition?
Select the correct answer below:

A) Hail is formed from water droplets lifted by air currents to an altitude where they turn into pellets of ice.

B) Frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

C) In the winter, the top few inches of a pond turn to ice.

D) The visible cloud arising from a boiling tea kettle is not actually steam, but droplets of liquid water that form as the
steam cools in the air.

Answers

Answer:

b

Explanation:

deposition is when water turns from gas to solid. b is the only one that fits

Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

What is deposition?

Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.

Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

Learn more about deposition here: https://brainly.com/question/6110024

1. A
a stiff structure that surrounds and
protects a coll: found in plant, fungus, and some bacteria cells.
2.
is living things consisting of many cells.
3.
a green pigment that traps energy
from the sun.
the process in which plants and
some other organisms use the energy in sunlight to make food.
5. A
found in the nucleus of a cell,
a long nucleic acid molecule containing the genetic instructions
for the development and functioning of all living organisms.

Answers

Answer:

1 cell wall

2 yes

3 chloroplast

4 photosynthesis

5 Deoxyribonucleic acid (I believe)

hope this helped a little and pls mark brainiest if it did :)

Explanation:

The cell wall is a rigid layer that is found outside the cell membrane and surrounds the cell, providing structural support and protection.

how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

Answers

Answer:

how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.

Answers

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

For more about electronic configuration, see:

https://brainly.com/question/4949433

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