Ba(OH)2:_______.
A. 1 barium atom, 1 oxygen atom and 1 hydrogen atom.
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
C. 1 barium atom, 2 oxygen atoms and 2 hydrogen atoms.
D. 1 barium atom, 2 oxygen atoms and 1 hydrogen atom.

Answers

Answer 1

Answer: D

Explanation: Expand this (OH)2 you will get 2O, 2H

Hence 1Ba, 2O, 2H

Answer 2

Answer:

B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.


Related Questions

Three different students determined the density of a metal object. Here are their results: 15.12 g/mL, 15.09 g/mL, and 15.12 g/mL. The actual density of the object was 14.41 g/mL. Calculate the percent error. Make sure to include units with your answer, units are %.

Answers

Answer:

The correct answers are 4.93 %, 4.72 % and 4.93 %.

Explanation:

Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,  

% error = (actual value - calculated value) / actual value * 100

By 1st student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41 - 15.12) / 14.41 * 100  

= 0.71/14.41 * 100

= 4.93 %

By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,  

% error = (14.41-15.09)/14.41 * 100

= 0.68/14.41 * 100

= 4.72 %

By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41-15.12)/14.41 * 100

= 0.71/14.41 * 100

= 4.93 %

3 Pieces of Key Information elements, compounds, mixtures also state a real world example of elements, compounds, mixtures and lastly why is elements, compounds, mixtures are important

Answers

Answer:

Element and compounds are the pure substances but mixture is not a pure substance.

Explanation:

Element and compounds are the pure substances in which element comprise of only one atom while compound is formed by the chemical combination of more than one element in a fixed ratio by mass while mixture is also made up of more that one substances but they are combine physically not chemically. Elements, compounds, mixtures are very important because all the materials we used in our daily life are formed from elements, compounds and mixtures.

What is the number of valence electrons in a nitrogen atom in the ground state

Answers

Answer: 5

Explanation:

It just is

Answer:

5

Explanation:

Bc valence electron means last # in the electron configuration

Compound H is optically active and has the molecular formula C6H10 and has a five carbon ring. On catalytic hydrogenation, H is converted to I (C6H12) and I is optically inactive. Propose structures for H and I. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

Answers

Answer:

Explanation:

Given that ;

Compound H is optically active and have a molecular formula of C6H10 and therefore undergo catalytic hydrogenation. Catalytic hydrogenation involves the use Platinum/Nickel to produce C6H12

i.e

[tex]C_6H_{10} +H_2 \to ^{Pt/Ni} \ \ \ C_6H_{12}[/tex]

The proposed H and I structures are shown in the diagrams attached below .

compound H represents  3- methyl cyclopentene

compound I represents methyl cyclopentane

However; 3- methyl cyclopentene posses just only one chiral carbon which is optically active at the third position and it R and S enantiomers are shown in the second diagram below.

The starting material is  3-methylcyclopentene while the optically inactive product is 1-methyl cyclopentane.

Hydrogenation refers to the addition of hydrogen across the double bond of an unsaturated compound. Hydrogenation results in the formation of a saturated compound having two more hydrogen atoms than the starting material.

The starting material is optically active 3-methylcyclopentene. The R and S enantiomers of the starting material is shown in image (1) attached. The optically inactive product is, 1-methyl cyclopentane is shown in image (2) attached.

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5. Rosalind Franklin was a key figure in the discovery of the structure of DNA, yet she
was not included in the Nobel Prize which was awarded to Watson and Crick. Carry out
some research to find out how she contributed to this work and use the space below
to write up your findings​

Answers

Answer:

Explanation:

Search for "Rosalind Franklin: DNA's unsung hero - Cláudio L. Guerra" which basically summarizes what Rosalind did and how we was snubbed from receiving the noble prize even though she had vast and critical evidence to highlight the structure of DNA. You can look for more sources but I can tell you a quick recap:

Rosalind Franklin was born in an era where women scientists or workers were very uncommon and they were even discriminated and looked down upon. After her phD., she was working to find the structure of DNA and soon she was able to form an x-ray image of it. However, her lab colleague took the picture and showed it to other scientists (Watson and Crick) without the knowledge or permission of Rosalind. Here Rosalind was working on analyzing her data and on other part of world Watson and Crick were doing the same. Based on Watson and Crick's analysis, they came up with the correct structure of DNA and soon Rosalind got done as well. Both submitted their paper to journal, however, the journal placed Watson and Crick paper before Rosalind (making it look like Rosalind just confirmed what Watson and Crick proposed). This made it look like Watson and Crick were geniuses behind DNA structure whereas, in reality, it was Rosalind. She would have received Nobel Prize but she died of Cancer and Nobel prizes are not awarded to dead people.

Which metal can replace another metal in a reaction

Answers

Answer:

The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)

Explanation:

The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)

12.39 g sample of phosphorus (30.97 g/mol) reacts with 52.54 g of chlorine gas, Cl2
(70.91 g/mol) to form only phosphorus trichloride, PC13 (137.33 g/mol). Which is the
limiting reactant?

Answers

Answer:

P is the limiting reagent

Explanation:

P = phosphorus  = 30.97g/mol

Cl2 = Chlorine = 70.91g/mol

PCl3 = Phosphorus Trichloride = 137.33g/mol

P + Cl2 = PCl3

Left Side

P = 1

Cl = 2

Right Side

P = 1

Cl = 3

So equation needs to be balanced first

2P + 3Cl = 2PCl3

Left Side

P = 2

Cl = 6

Right Side

P = 2

Cl = 6

That's better.

Ok so we have 12.39g of P so we have 0.4 moles of it

We then have 52.54g of Cl so we have 0.74 moles of it

For every P we need 1.5 Cl so we have an excess of Cl

need help and quick answer as fast as possible

Answers

yes. arthropod are animals such as insects, crabs, lobsters etc

A student has an unknown sample of solution X. This solution is placed in a 1.00 cm wide cuvet and inserted into the spectrometer, producing an absorbance reading of 0.275 at a wavelength of 525.0 nm. What is the concentration of solution X in the unknown sample

Answers

Answer:

The concentration of the sample is 3.564x10⁻³M.

Explanation:

Using Lambert-Beer law, absorbance of a sample is directely proportional to its concentration.

The general graph of the absorbance of the standards with different concentrations is:

Y = 75.9X + 0.0045

R² = 0.9946

Where Y is the absorbance of the sample and X its concentration in mole/L.

If a solution has an absorbance of 0.275:

0.275 = 75.9X + 0.0045

0.2705 = 75.9X

3.564x10⁻³M = X → The concentration of the sample.

Classify the following unbalanced chemical reaction Na3PO4(aq) + FecCl2(aq) = Fe3(PO4)2(s) + NaCl(aq)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction

Answers

Answer:

2. Precipitation Reaction

Explanation:

Na3PO4(aq) + FecCl2(aq) = Fe3(PO4)2(s) + NaCl(aq)

Fe3(PO4)2(s) - solid, it means it will precipitate.

It is a precipitation reaction.

What is precipitate give example?

A precipitate is a solid that forms out of a solution. A common example is that of the mixing of two clear solutions: (1) silver nitrate (AgNO3) and (2) sodium chloride (NaCl): The reaction is. The precipitate forms because the solid (AgCl) is insoluble in water.

What is precipitate formation?

A precipitate is a solid formed in a chemical reaction that is different from either of the reactants. This can occur when solutions containing ionic compounds are mixed and an insoluble product is formed. The identity of the precipitate can often be determined by examining solubility rules.

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Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answers

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

A sample of chemical X is found to contain 5.0 grams of oxygen, 10.0 grams of carbon, and 20.0 grams of nitrogen. The law of definite proportion would predict that a 75 gram sample of chemical X should contain how many grams of carbon

Answers

Answer: 75 grams sample of chemical X should contain 21.43 grams of carbon

Explanation: The law of definite proportion states that a given chemical compound always contains its component elements in fixed ratio.

From the question, chemical X contains 5.0 grams of oxygen, 10.0 grams of carbon, and 20.0 grams of nitrogen.

Sum up the masses

5.0g + 10.0g + 20.0g = 35.0g

This means, 10.0 grams of carbon are present in 35.0 grams of chemical X.

Now, to the determine the mass of carbon that 75 gram sample of chemical X should contain,

According to the law of definite proportion, the component elements of a given chemical compound are in fixed ratio. Therefore,

If 35.0g of chemical X contains 10.0g of carbon

Then, 75 g of chemical X will contain

(75 × 10) / 35 g

= 21.43 grams

Hence, 75 grams sample of chemical X should contain 21.43 grams of carbon.

Answer:

According to the law of definite proportion, a 75 gram sample of chemical X should contain 21.249 grams of carbon.

Explanation:

The total mass of the sample is equal to the sum of masses of oxygen, carbon and nitrogen. That is:

[tex]m_{tot} = m_{O} + m_{C} + m_{N}[/tex]

If [tex]m_{O} = 5\,g[/tex], [tex]m_{C} = 10\,g[/tex] and [tex]m_{N} = 20\,g[/tex], then:

[tex]m_{tot} = 35\,g[/tex]

According to the law of definite proportion, the following simple rule of three is used:

[tex]m_{C'} = m_{C} \times \frac{m_{tot'}}{m_{tot}}[/tex]

If [tex]m_{C} = 10\,g[/tex], [tex]m_{tot} = 35\,g[/tex] and [tex]m_{tot'} = 75\,g[/tex], then:

[tex]m_{C'} = 10\,g\times \frac{75\,g}{35\,g}[/tex]

[tex]m_{C'} = 21.429\,g[/tex]

According to the law of definite proportion, a 75 gram sample of chemical X should contain 21.249 grams of carbon.

Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?

Answers

Answer:

ΔH = - 44.0kJ

Explanation:

H2O(l)→H2O(g), ΔH =44.0kJ

In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH  is positive.

If he reaction is reversed, we have;

H2O(g)→H2O(l)

In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still  have the same value.

Covalent bonds can be best described as

Answers

Answer:

neutral atoms coming together to share electrons

Answer:

a

Explanation:

neutral atoms coming together to share electrons

In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

Answers

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]

[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]

so,

[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable  until q = Keq

If a gas occupies 12.60 liters at a pressure of 1.50 atm, what will its pressure at a volume of 2.50 liters?

Answers

Answer:

7.56 atm

Explanation:

Boyle's law states that the pressure and volume of a gas are proportional to each other

The formular for Boyle's law is

P1V1=P2V2

According to the question above, the values given are

P1=1.50 atm

P2= ?

V1=12.60 litres

V2= 2.50 litres

Let us make P2 the subject of formular

P2= P1V1/V2

P2= 1.50×12.60/2.50

P2= 18.9/2.50

P2= 7.56 atm

Hence when the volume of a gas is 2.50 litres then it's pressure is 7.56 atm

At 298K, the equilibrium constant for the following reaction is 4.20×10-7: H2CO3(aq) + H2O H3O+(aq) + HCO3-(aq) The equilibrium constant for a second reaction is 4.80×10-11: HCO3-(aq) + H2O H3O+(aq) + CO32-(aq) Use this information to determine the equilibrium constant for the reaction: H2CO3(aq) + 2H2O 2H3O+(aq) + CO32-(aq)

Answers

Answer:

The correct answer is 2.016 x 10⁻¹⁷

Explanation:

We have the following chemical reactions and their equilibrium constants (K):

(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)  K₁= 4.20×10⁻⁷

(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)    K₂= 4.80×10⁻¹¹

And we have to obtain K for the following reaction:

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)

If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.

H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)     K₁= 4.20×10⁻⁷

+

HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)      K₂= 4.80×10⁻¹¹

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

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)    K= K₁ x K₂

K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷


Discuss any give ways by which
the falling moral standards of Ghanaian
youth can be minimised.

Answers

Answer:

The falling standards of Ghanaian youths can be minimized by proper upbringing of the children by their parents. The youths should be taught about what is wrong or right and there should be a corresponding reward for those who do good and exceptional in order to encourage others in towing that line and punishment should also be meted out to those who break the law. Mediocrity shouldn’t be celebrated and the elders should lead by example.

These will make the falling standards of Ghanaian youth get reduced.

10. How many grams of NH, are present in 6 moles
of NH,?​

Answers

Answer:

90.08784 grams

Explanation:

idk

Classify the following unbalanced chemical reaction Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction

Answers

Answer:

3. Oxidation-Reduction Reaction

Explanation:

Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)

Fe(s) -2e- ----> Fe2+(aq)  oxidation

Cl2(aq)  + 2e- -----> 2Cl-(aq)  reduction

The given unbalanced chemical reaction is the oxidation-reduction reaction. Therefore, option (3) is correct.

What is an oxidation-reduction reaction?

Redox reactions can be defined as oxidation-reduction chemical reactions in which the reactants of the reaction undergo a change in their oxidation states. All the redox reactions are further broken down into two different processes: a reduction process and an oxidation process.

The oxidation and reduction reactions take place simultaneously in an Oxidation-Reduction reaction. The substance that is getting reduced in a reaction is known as the oxidizing agent, while a substance that is getting oxidized is the reducing agent.

The given chemical reaction is:

[tex]Fe(s) + Cl_2(aq) \longrightarrow Fe^{2+}(aq) + Cl^-(aq)[/tex]

The oxidation reaction for this reaction is: Fe (s)  →  Fe²⁺ (aq)  + 2e⁻

The reduction reaction:  Cl₂ (g)  +  2e⁻   →  2Cl⁻ (aq)

Therefore, the given reaction between the iron and chlorine gas is the oxidation-reduction reaction or redox reaction.

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Take a series of observations to determine if process is spontaneous. Based upon those observations, you will create an activity series, listing the metals in order of their reactivity. Second, you will construct a series of virtual galvanic cells and use those to power a stopwatch. Third, you will determine the standard reduction potential of an unknown metal; comparing its reduction potential to a standard list, you will identify the unknown. Finally, you will create a situation in which the cells are not in the standard condition and measure the cell potential; using the Nernst equation, you will determine the concentration of an unknown solution
Answer the below questions for the portion of the activity in which Sn(s) is placed in AgNO3(aq)
1. Is there a reaction? (circle the correct response) Yes / No
2. How many electrons are transferred 4 electrons
3. Write the balanced redox reaction for the combination of AgNO3(aq) and Sn(s)Sn(s) + Ag+(aq)  Sn2+(aq) + Ag(s)

Answers

Answer:

Explanation:

2AgNO₃ + Sn ⇄ Sn( NO₃)₂ + 2Ag

Ag⁺/Ag = .80 V

Sn⁺²/Sn = - .14 V

Hence Ag will be reduced and Sn will be oxidised . Hence the reaction will take place . YES .

2 ) 2 electrons are transferred .

3 )

2Ag⁺  + 2e = 2Ag

Sn = Sn⁺²  + 2e

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

2Ag⁺ + Sn = Sn⁺²  + 2Ag .

When a sample of Mg(s) reacts completely with O2(g), the Mg(s) loses 5.0 moles of electrons. How many moles of electrons are gained by the O2(g)? *

Answers

Answer:

if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!

Explanation:

Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.

Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.

Oxidation half equation;

Mg(s)-----> Mg^2+(aq) + 2e

Reduction half equation;

O2(g) + 2e ------> 2O^2-(aq)

From the balanced reaction equation, two moles of electrons is transferred.

Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.

Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm

1. Calculate the volume of the bar:

2. Calculate the (experimental) density of the bar:

3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?

4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%

Answers

Answer:

1= Volume

= Length x breath x height

= 13.90 x 2.9 x 0.081

=3.26511

2= Density = Mass ÷ volume

= 11.3 ÷ 3.26511

= 3.461 (3d.p)

idk the rest because you haven't shown a picture of the rest

Answer:

1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%

Explanation:

Experimental data:

Mass          = 11.3    g

Length      = 13.90 cm

Width        =  2.9    cm

Thickness = 0.081 cm

Calculations:

1. Volume of bar

V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³

2. Experimental density

[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]

3. Identity of metal

The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)

The metal is probably barium.

4. Percent difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]

Which compound has the lowest melting point? KCl CaCl2 Na2O C6H12O6

Answers

It is called ethane.

If a jet’s cruising altitude is 32,200ft(to three significant figures),the distance in km is :(1 mile=1.61km;1 mile=5280 ft)

Answers

Answer:

9.82 km.

Explanation:

Hello,

In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:

[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]

Best regards.

A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answers

Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

Explanation:

Formula used for Elevation in boiling point :

[tex]\Delta T_b=k_b\times m[/tex]

or,

[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]

[tex]k_b[/tex] = boiling point constant  = ?

m = molality

[tex]w_2[/tex] = mass of solute (urea) = 55.4 g

[tex]w_1[/tex] = mass of solvent  X =  500 g

[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]

[tex]k_b=2.4^0C/m[/tex]

Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW

Answers

Answer:

Explanation:

From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer

0.25 L elution buffer = 250 mL elution butter

The breaking buffer that we use this week contains

10mM Tris    =   0.01 M

150mM NaCl  =   0.15 M

300mM imidazole.  = 0.3 M

The stock concentration  of Tris in 1M

Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;

[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 2.5 \ mL[/tex]

In NaCl, The amount of stock concentration is 5 M

so; using the same formula; we have:

[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]

[tex]V_1 = 7.5 \ mL[/tex]

From Imidazole ; the amount of stock concentration is

[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 75 \ mL[/tex]

Thus; we can have a table as shown as :

Stock concentration        volume to be added        Final concentration

1 M of Tris                              2.5 mL                            10 mM

5 M of  NaCl                          7.5 mL                             150 mM

1 M of Imidazole                    75  mL                            300  mM

In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.

What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?

Answers

Answer:

Al2(SO4)3 is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ba + Al2(SO4)3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2(SO4)3.

From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.

Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.

One method of experimentally determining whether a species is paramagnetic is to weigh it in an instrument called a magnetic susceptibility balance. This is a balance with a strong electromagnet placed next to the sample holder. If the species is paramagnetic, the mass reading of the balance will increase when the field is switched on.Classify these species as paramagnetic or diamagneticWhich species will have the strongest mass shift on a magnetic susceptibility balance?

Answers

Answer:

Diamagnetic have paired electrons while paramagnetic have at least one unpaired electron.

Explanation:

F2, C2 and N2 are diamagnetic while O2 and B2 are paramagnetic. Diamagnetic are those atoms which have paired electrons while paramagnetic are those atoms which contain at least one unpaired electron so we can say that F2, C2 and N2 have paired electrons while O2 and B2 have unpaired electrons. When diamagnetic materials are allowed to contact with external magnetic field so they will be repelled while paramagnetic materials are attracted by magnetic field due to the presence of unpaired electrons.

Iron oxide (FeO) is the strongest paramagnetic material having the value of 720.

1- A volumen constante un gas ejerce una presión de 880 mmHg a 20º Celsius dentro de una olla a presión ¿Qué temperatura habrá si el marcador de presión muestra un valor de 1050 mmHg?

Answers

Answer:

In this problem the correct thing would be to use the ideal gas equation.

Explanation:

Well in this exercise we will use the following equation:

(P x V) / T = (p x v) / t

On the right side of the equation we will find the initial values, that is, the values ​​with which the reaction begins and in general they are always the first to write in the problems.

Instead on the left side of the equation, the letters that are in lowercase are the final values, that is to say at the end of the reaction that the values ​​of pressure, temperature and volume are reached.

P is pressing, just like p, T and t are temperature, and V and v are volume.

We use this equation so we consider the behavior of said gas to be an IDEAL gas, a constant volume.

That is why the given pressures require an atmosphere to pass, which is another unit used to press the pressure ... Much needed in this equation! An atmosphere is equivalent to 760 millimeters of mercury ...

Then the final and initial pressures would be:

initial pressure: 1.15 atm

final pressure: 1.38 atm

In this way you already have the values ​​to be able to solve in the equation your unknown that would be the final temperature:

Considering that the volume is constant, we cancel it from the equation, 1.15 atm would be in the value of P and 1.38 in the value of p ... In this way it considers that 20 degrees Celsius is the initial temperature or ses T, we would only have to clear the t.

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