An 80L capacity steel cylinder contains H2 at a pressure of 110 atm and 30 ° C, after extracting a certain amount of gas, the pressure is 80 atm at the same temperature. How many liters of hydrogen (measured under normal conditions) have been extracted?

Answers

Answer 1

Answer:

2200 L

Explanation:

Ideal gas law:

PV = nRT,

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The initial number of moles is:

(110 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K

n = 353.58 mol

After some gas is removed, the number of moles remaining is:

(80 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K

n = 257.15 mol

The amount of gas removed is therefore:

n = 353.58 mol − 257.15 mol

n = 92.43 mol

At normal conditions, the volume of this gas is:

PV = nRT

(1 atm) V = (92.43 mol) (0.0821 L atm / K / mol) (273.15 K)

V = 2162.5 L

Rounded, the volume is approximately 2200 liters.


Related Questions

A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.

Answers

Answer:

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume of the unit cell

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

7.38 g/cm³ is the density of the metal

The density of the metal is 7.40 g/cm³

In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.

It is expressed by using the formula:

[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]

where;

[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge length

For face-centered cubic FFC;

The edge length  [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]

[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]

[tex]\mathbf{a =560.0285 \ pm }[/tex]

a = 5.60 × 10⁻⁸ cm

Replacing it into the previous equation, we have:

[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]

[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]

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How many unit cells share an atom that is located at the center of a cube edge of a unit cell?

Answers

Answer:

zero

Explanation:

In a unit cell, an atom that is located at the center of a cube edge is not involved in sharing unit cells because a central atom of a unit cell belongs to the entire cell and only to that unit cell of the lattice.

Hence, the center atom of a unit cell do not share any unit cell and the correct answer is "Zero".

H2S(g) 2H2O(l)3H2(g) SO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.60 moles of H2S(g) react at standard conditions.

Answers

Answer: [tex]\Delta S[/tex] = 473.92J/K.mol

Explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:

[tex]\Delta S = m\Sigma product - n\Sigma reagent[/tex]

The balanced reaction:

[tex]H_{2}S_{(g)}+2H_{2}O_{(l)}=>3H_{2}_{(g)}+SO_{2}_{(g)}[/tex]

gives the proportion reagents react to form products, so, if 1.6 moles of [tex]H_{2}S_{(g)}[/tex]:

3.2 moles of water is used;

4.8 moles of hydrogen gas is formed;

1.6 moles of sulfur dioxide is also formed;

Calculating entropy change:

[tex]\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)[/tex]

[tex]\Delta S=628.8+398.08-328.96-224[/tex]

[tex]\Delta S[/tex] = 473.92J/K.mol

Entropy change for the given chemical reaction is [tex]\Delta S[/tex] = 473.92J/K.mol

Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.

Answers

Answer:

pH = 3.49

Explanation:

We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH  of a buffer ssytem using the Henderson-Hasselbach equation.

pH = pKa + log [base] / [acid]

pH = -log Ka + log [NO₂⁻] / [HNO₂]

pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M

pH = 3.49

The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49

We'll begin by calculating the the pKa of acid. This can be obtained as follow:

Acid dissociation constant (Ka) = 4.50×10¯⁴

pKa =?

pKa = –Log Ka

pKa = –Log 4.50×10¯⁴

pKa = 3.35

Finally, we shall determine the pH of the solution.

pKa = 3.35

Concentration of HNO₂, [HNO₂] = 0.210 M

Concentration of KNO₂, [KNO₂] = 0.290 M

pH =?

The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:

pH = pKa + log [base] / [acid]

pH = pKa+ log [NO₂⁻] / [HNO₂]

pH = 3.35 + log (0.290 / 0.210)

pH = 3.49

Thus, the pH of the solution is 3.49

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Complete the unit conversion by entering the correct numbers
A=
B=
C=

Answers

Answer: A=1, B=3, C=12

Explanation:

For this problem, you will need to know your unit conversions. There are 3 ft in 1 yard. Knowing this, we can find A, B, C.

For A and B, we know that we want to cancel out ft so the answer can be in yards. To do so, we need to put B=3 and A=1.

Now that we know the unit conversion, we can directly solve.

36 ft×(1 yd/3 ft)=12 yd

Our final answer is A=1, B=3, C=12.

Answer:

A=1,000, B=1, C=5,400

Explanation:

the question was 5.4L x AmL / BmL = CmL

4. Given that the enthalpy of reaction for a system at 298 K is -292 kJ/mol and the entropy for that system is 224 J/mol K, what's the free energy for the system?
A.-87,793 kJ
B.-358 kJ
C.-225 kJ
D. -66,751 kJ​

Answers

Answer:

[tex]\Delta G=-359\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:

[tex]\Delta G=\Delta H -T\Delta S\\[/tex]

In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:

[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]

Best regards.

Answer:

B- 358 kj

Explanation: I took the test

What is the molecular mass of this substance? a. 31.02 u b. 63.02 u c. 47.02 u d. 126.04 u e. 110.01 u

Answers

Answer:

Molecular mass of HNO₃ = 63.015 g/mol

Explanation:

Note: The given question is incomplete , the given substance is nitric acid

Given:

Nitric acid (HNO₃)

Find:

Molecular mass

Computation:

Molecular mass of HNO₃ = 1.008 + 14.007 + 3(16)

Molecular mass of HNO₃ = 1.008 + 14.007 + 48

Molecular mass of HNO₃ = 63.015 g/mol

Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?

Answers

Answer:

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

Rate constant:

t(1/2) = ln 2 / K

As half-life of Cesium-137 is 30.2 years:

30.2 years = ln 2 / K

K = 0.02295 years⁻¹

Replacing this result and with the given data of the problem:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.02295 years⁻¹* t + ln[A]₀

Ln ([A] / [A₀]) = -0.02295 years⁻¹* t

As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:

Ln (0.2) = -0.02295 years⁻¹* t

70.1 years = t

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms

Answers

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

What type of chemist exclusively studies most carbon compounds?
-biochemist
-physical chemist
-organic chemist
-inorganic chemist

Answers

Answer:

Organic chemist? I do not know.

Explanation:

Thanks you.

The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.

What is an organic chemist ?

The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.

Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.

Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.

According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.

Thus, option C is correct.

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What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)

Answers

Answer: d=2000 g/L

Explanation:

Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.

[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]

Now that we have grams, we can divide to get density.

[tex]d=\frac{10000g}{5 L}[/tex]

d=2000g/L

A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was ________ mm Hg.

Answers

Answer:

96 mmHg

[tex]h=96mmHg[/tex]

Explanation:

From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .

The difference of height of the mercury level is given as,

h=9.60cm

h=9.60(10mm/1cm)

[tex]h=96mm[/tex]

But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height

P(gas)=h

P(gas)=96mmHg

This pressure is as a result of the presence of gas.

Therefore, the pressure of the argon gas in the container is 96mmHg.

The pressure of the argon in the container was 96mmHg.

We were told that the manometer has closed ends which means that the

height will not be affected by atmospheric pressure.

The height which is the difference in mercury level is

h=9.60cm

We can convert it to millimeter by multiplying it by 10

h=9.60 × 10 = 96mm

The pressure of the closed end manometer will be equal to the height

P(gas)=h

P(gas)=96mmHg

The pressure of the argon gas in the container is 96mmHg.

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If we represent the equilibrium as:...N2O4(g) 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature is increased the equilibrium concentration of NO2: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.

Answers

Answer:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases.

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Explanation:

Hello,

In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.

With the aforementioned, we can conclude that the chemical reaction:

[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]

Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:

1. This reaction is: B. Endothermic.

2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:

[tex]G=H-TS[/tex]

As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:

[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]

3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.

Regards.

PLEASE HELP!! 40 POINTS

Answers

Answer:

1) 6.524779402×10^(-17)  

2)521.1g

3)113

Explanation:

Answer: 1) 6.524779402×10^(-17)

2)521.1g

Explanation:

3. What is the mass of an object with a volume of 4 L and a density of 1.25 g/mL?

Answers

Answer:

5000g

Explanation:

mass= density × volume

Since the unit of density here is g/mL, we need to convert the volume to mL.

1L= 1000mL

4L= 4 ×1000 = 4000 mL

Mass of object

= 1.25 ×4000

= 5000g

Answer:

5,000 grams

Explanation:

The mass of an object can be found by multiplying the volume by the density.

mass= volume * density

The density is 1.25 g/mL and the volume is 4 L.

First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.

There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.

4 * 1,000 = 4,000

The volume is 4,000 mL.

Now, find the mass of the object.

mass= volume * density

volume = 4,000

density= 1.25

mass= 4,000 * 1.25 = 5,000

Add the appropriate units for mass, in this case, grams, or g.

mass= 5,000 g

The mass of the object is 5,000 grams.

For the following reaction, predict whether the equilibrium lies predominantly to the left or to the right. Explain.
NH4+(aq) + Br-(aq) - NH3(aq) + HBr(aq)
If Ka
(NH4+) = 5.6 x 10-10

Answers

Answer:

Lies predominantly to the left.

Explanation:

In the reaction:

NH4⁺(aq) + Br-(aq) ⇄ NH3(aq) + HBr(aq)

Conjugate acid + Ion ⇄ weak base + strong acid

HBr is a strong acid whereas NH3/NH4⁺ are the weak base and its conjugate base. A strong acid as HBr dissociates completely in solution as H⁺ and Br⁻. That means in solution you will never have HBr without dissociation doing the reaction:

lies predominantly to the left.

How many grams is 5.8 moles of hydrochloric acid (HCI)?
Answer to the nearest 0.01 g.

Answers

Answer:

211.47 grams

Explanation:

We need to set up a dimensional analysis to solve this problem by converting from moles to grams.

First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:

1.01 + 35.45 = 36.46 g/mol

We have 5.8 moles of HCl, so multiply by its molar mass:

(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g

The answer is thus 211.47 grams.

~ an aesthetics over

Answer:

[tex]\large\boxed{211.47}\\[/tex] grams

Explanation:

First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.

Hydrogen has an atomic weight of 1.00784 and Chlorine has an atomic mass of 35.453.Add those two values together - 1.00784 + 35.453 = 36.46084Multiply this value by 5.8 (one mole is equivalent to the atomic mass of the compound) - 5.8 x 36.46084 = 211.472872Round to the nearest 0.01 gram - 211.47

[tex]\large\boxed{211.47}[/tex] is the final answer.

How many moles of barium sulfate are produced from 0.100 mole of barium chloride?

Answers

Answer:

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Explanation:

Barium chloride and sodium sulfate react according to the following balanced reaction:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and products participate in the reaction:

BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl : 2 moles

Then you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?

[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]

amount of moles of BaSO₄= 0.100

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Human blood typically contains 1.04 kg/L of platelets. A 1.89 pints of blood would contain what mass (in grams) of platelets

Answers

A 1.89 pints of blood would contain 873 grams of platelets.

To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:

1 pint = 473.2 mL

                                  [tex]1.89 \times 473.2 = 894.3 mL[/tex]

1000 L = 1mL

         

                                         [tex]\frac{894.3}{1000}= 0.84L[/tex]

Now, just calculate the amount of platelets present in 0.84L:

                                    [tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]

                               

                                       x = 873 grams

So, a 1.89 pints of blood would contain 873 grams of platelets.

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1. What volume in milliliters of 0.100 M HClO₃ is required to neutralize 40.0 mL of 0.140 M KOH? 2. A 25.0 mL solution of HNO₃ is neutralized with 15.7 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?

Answers

Answer:

The correct answer is 1) 56 ml and 2) 0.314 M

Explanation:

1. The reaction taking place in the given case is,  

HClO₃ + KOH ⇒ KClO₃ + H2O, the molarity of HClO₃ given is 0.100 M, the molarity of KOH given is 0.140 M and the volume of KOH given is 40 ml, there is a need to find the volume of HClO₃.  

Therefore, the mole of HClO₃ = mole of KOH

= MHClO₃ × VHClO₃ = MKOH × VKOH

= 0.100 M × VHClO₃ = 0.140 M × 40 ml

VHClO₃ = 0.140 M × 40 ml/0.100 M

VHClO₃ = 56 ml.  

2. The reaction taking place is,  

2HNO₃ + Ba(OH)₂ ⇒ Ba(NO₃)₂ + 2H₂O

The volume of HNO₃ given is 25 ml, the molarity of Ba(OH)2 is 0.250 M, the volume of Ba(OH)2 is 15.7 ml, the n or the number of moles of HNO₃ is 2, and the n of Ba(OH)2 is 1, the concentration or M of HNO₃ is,  

M₁V₁/n₁ = M₂V₂/n₂

M₁ × 25/ 2 = 0.25 × 15.7/1

M₁ or molarity of HNO₃ = 0.314 M

1. The volume of HClO₃ required to neutralize the KOH is 56.0 mL

2. The concentration of the original HNO₃ solution is 0.314 M

1.

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

HClO₃ + KOH → KClO₃ + H₂O

This means,

1 mole of HClO₃ is required to neutralize 1 mole of KOH

From the titration formula

[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]

Where

[tex]C_{A}[/tex] is the concentration of acid

[tex]C_{B}[/tex] is the concentration of base

[tex]V_{A}[/tex] is the volume of acid

[tex]V_{B}[/tex] is the volume of base

[tex]n_{A}[/tex] is the mole ratio of acid

[tex]n_{B}[/tex] is the mole ratio of base

From the given information,

[tex]C_{A} = 0.100 \ M[/tex]

[tex]C_{B} = 0.140 \ M[/tex]

[tex]V_{B} = 40.0 \ mL[/tex]

From the balanced chemical equation

[tex]n_{A} = 1[/tex]

[tex]n_{B} =1[/tex]

Putting the values into the formula, we get

[tex]\frac{0.100 \times V_{A} }{0.140 \times 40.0} = \frac{1}{1}[/tex]

∴ [tex]0.100 \times V_{A} = 0.140 \times 40.0[/tex]

[tex]V_{A}=\frac{0.140\times 40.0}{0.100}[/tex]

[tex]V_{A}=\frac{5.60}{0.100}[/tex]

[tex]V_{A}=56.0 \ mL[/tex]

Hence, the volume of HClO₃ required to neutralize the KOH is 56.0 mL

2.

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O

This means, 2 mole of HNO₃ is required to neutralize 1 mole Ba(OH)₂  

From the given information,

[tex]V_{A} = 25.0\ mL[/tex]

[tex]C_{B} = 0.250 \ M[/tex]

[tex]V_{B} = 15.7 \ mL[/tex]

From the balanced chemical equation

[tex]n_{A} = 2[/tex]

[tex]n_{B} =1[/tex]

Also, Using the titration formula

[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]

We get

[tex]\frac{C_{A} \times 25.0 }{0.250 \times 15.7} = \frac{2}{1}[/tex]

Then,

[tex]C_{A} = \frac{2\times 0.250 \times 15.7} {1 \times 25.0}[/tex]

[tex]C_{A} =\frac{7.85}{25.0}[/tex]

[tex]C_{A} =0.314 \ M[/tex]

Hence, the concentration of the original HNO₃ solution is 0.314 M

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Chemistry
What is a chemical reaction

Answers

Answer:

A process that involves rearrangement

Explanation:

A chemical reaction is the process that involves rearrangement of the molecular or ironic structure of a substance, as a distinct from a change in physical form or a nuclear reaction.

Answer:

Explanation:

Chemistry

The chemical reaction H2(g) + ½ O2(g) → H2O(l) describes the formation of water from its elements.

The reaction between iron and sulfur to form iron(II) sulfide is another chemical reaction, represented by the chemical equation:

8 Fe + S8 → 8 FeS

Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Answers

Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

Arrange the following substances in the order of increasing entropy at 25°C. HF(g), NaF(s), SiF 4(g), SiH 4(g), Al(s) lowest → highest

Answers

Answer:

Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)

Explanation:

Hello,

In this case, we can arrange the increasing order of entropy at 25 \°C by taking into account, at first, that since solids are more molecularly organized than gases, the first we have solid sodium fluoride and solid aluminium, but in this case, as the higher the molar mass, the higher the entropy, the molar mass of aluminium is 27 g/mol and 42 g/mol for sodium fluoride, therefore, we first have:

Al(s)<NaF(s)

Afterwards, since the molar mass of hydrogen fluoride (HF), silicon fluoride (SiF4) and silane (SiH4) are 20, 104 and 32 g/mol respctively, since silicon fluoride has the greater molar mass, it also has the higher entropy. In such a way, the overall order turns out:

Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)

Best regards.

Given the following values of pKa, determine which is the weakest base of the answers listed. Acid pKa HClO2 1.95 HClO 7.54 HCOOH 3.74 HF 3.17 HNO2 3.15

Answers

Answer:

HClO 7.54

Explanation:

Hypochlorous acid (HClO) is a weakest acid because the pKa value of Hypochlorous acid is very high among the options given in the activity. pKa is a method which is used in order to identify the strength of an acid. The higher the value of pKa of a liquid, lower the strength of an acid while lower the value of pKa of chemical, higher the strength of an acid. In the options, HClO2  is a strong acid due to high lower pKa value.

Cite examples of how copper deposits occur. Choose one or more: A. as an agglomeration metal B. as a native metal C. in carbonate ore minerals D. in sulfide ore minerals

Answers

Answer:

A. as an agglomeration metal

B. as a native metal

D. in sulfide ore minerals

Explanation:

Copper is a metal with symbol Cu and atomic number 29. It has a pinkish-orange color and is malleable, ductile and has a high thermal and electrical conductivity. This is why it is often used in electrical appliances.

Copper exists as an agglomeration metal, as a native metal or in sulfide ore minerals such as Cu2S.

The examples of copper deposits are  agglomeration metal, as a native metal or in sulfide ore minerals. Option A, B, and D are correct.

 

Copper is a metal with high thermal and electrical conductivity. hence, it is often used in electrical appliances.

Copper found as an agglomeration metal, as a native metal or in sulfide ore minerals such as [tex]\bold { Cu_2S.}[/tex]

Therefore, the examples of copper deposits are  agglomeration metal, as a native metal or in sulfide ore minerals. Option A, B, and D are correct.

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how can you prevent frequent landslides from heavy rains

Answers

Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.

Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud

B. Plant vegetation on the slops that expreience landslides.

E. Reduce the sloped where landslides occur.

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An unknown gas diffuses 5 times slower than that of H2.The moleculer mass of unknown gas is??

Answers

Answer:

50.

Explanation:

We can write Graham's Law of Diffusion as:

(Rate 1)^2 = Molecular Mass 2

--------------    -------------------------

(Rate 2)^2    Molecular Mass 1

So using the Given Information:

1^2  / (1/5)^2 = Molecular Mass of unknown gas / 2, so:

25 = M/2

M = 50.

Which molecule is NOT hypervalent?
Select the correct answer below:
SF
PBr3
PBr5
XeFo

Answers

Answer:

PBr3 is NOT hypervalent

Explanation:

The molecule that is not hypervalent is PBr3

A molecule can be defined as the smallest part of a substance that can exist independently.

It is formed by the chemical combination of two or more atoms.

A molecule is said to be hypervalent when more than four pairs of electrons are around the central atom.

A molecule is said to be hypovalent when less than four pairs of electrons are around the central atom.

From the question, the molecule that is hypovalent is PBr3

This is because, phosphorus can make hypervalent compounds, but in this specific example it is sharing three bonds and has one lone pair, so it has simply a full octet.

Therefore, the molecule that is not hypervalent is PBr3.

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What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH

Answers

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630

Define the following terms - you may need to consult your lecture text or other suitable resource:
a. monomer,
b. repeating unit,
c. condensation polymerization,
d. cross-linked polymer

Answers

Answer:

a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.

b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.

c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.

d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers.  The cross-links leads to an alteration in the physical properties of the polymer.

The definition of following terms are :

a) Monomers:

The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.

The polymer shaped can too be in a three dimensional arrange.

b) Repeating unit:

A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.

A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.

c) Condensation polymerization:

This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.

d) Cross-linked polymer:

This can be a polymer shaped from a sort of holding of particles.

The cross-links leads to an modification within the physical properties.

Definitions

Definition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.

Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.

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