If 8.23 g of magnesium chloride react completely with sodium phosphate, how many grams of magnesium phosphateare produced

Answers

Answer 1

Answer:

The correct answer is 7.57 grams of magnesium phosphate.

Explanation:

Based on the given question, the chemical reaction taking place is:  

2Na₃PO₄ (aq) + 3MgCl₂ (aq) ⇒ Mg₃(PO₄)2 (s) + 6NaCl (aq)

From the given reaction, it is evident that two moles of sodium phosphate reacts with three moles of magnesium chloride to produce one mole of magnesium phosphate.  

Based on the given information, 8.23 grams of magnesium chloride reacts completely with sodium phosphate, therefore, magnesium chloride in the given case is the limiting reagent.  

In the given case, 3 moles of magnesium chloride produce 1 mole of magnesium phosphate. Therefore, 1 mole of magnesium chloride will produce 1/3 mole of magnesium phosphate.  

The molecular mass of magnesium chloride is 95.21 grams per mole. So, 1 mole of magnesium chloride is equivalent to 95.21 grams of magnesium chloride.  

On the other hand, the molecular mass of magnesium phosphate is 262.85 grams per mole. Therefore, 1 mole of magnesium phosphate is equal to 262.85 grams of magnesium phosphate.  

As seen earlier that 1 mole of magnesium chloride = 1/3 moles of magnesium phosphate. So,  

95.21 grams of magnesium chloride = 1/3 × 262.85 grams of magnesium phosphate

= 262.85 / 3 grams of magnesium phosphate

1 gram of magnesium chloride = 262.85 / 3 × 95.21 grams of magnesium phosphate

8.23 grams of magnesium chloride = 262.85 / 3 × 95.21 × 8.23 grams of magnesium phosphate

= 7.57 grams of magnesium phosphate

Hence, when 8.23 grams of magnesium chloride when reacts completely with sodium phosphate, it produces 7.57 grams of magnesium phosphate.  


Related Questions

Given the specific heat for aluminum is 0.900 J/g.°C, how much heat is released when a 3.8 g sample
of Al cools from 450.0°C to 25°C?
A. 1.5 kJ
B. 54 J
C. 60J
D. 1.7 kJ
E. 86 J

Answers

Answer:

Q = 1.5 kJ

Explanation:

It is given that,

The specific heat for aluminum is 0.900 J/g°C

Mass of sample, m = 3.8 g

Initial temperature, [tex]T_i=450^{\circ} C[/tex]

Final temperature, [tex]T_f=25^{\circ} C[/tex]

We need to find the heat released. The amount of heat released is given by the formula:

[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=3.8\times 0.9\times (25-450)\\\\Q=1453.5\ J\\\\Q=1.45\ kJ[/tex]

or

[tex]Q=1.5\ kJ[/tex]

So, the correct option is (A) i.e. 1.5 kJ.

Calculate the combustion of gaseous dimethyl ether CH 3 OCH 3 (g)+3O 2 (g) 2CO 2 (g)+3H 2 O(l) using standard molar enthalpies of formation Molecule AH H l ^ 0 (k)/mol) CH 3 OCH 3 (g) - 184.1

Answers

Answer:

[tex]\Delta _cH=-1328.3kJ/mol[/tex]

Explanation:

Helllo,

In this case, for the given chemical reaction in gaseous state:

[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]

We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:

[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]

Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:

[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]

Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.

Regards.

What is the absolute magnitude of the rate of change for [NH3] if the
rate of change for [Hz] is 9.00 M/s in the reaction 2 NH3(g) → N2(g) +
3 H2(g)?

Answers

Answer:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:

[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]

In such a way, knowing the rate of formation hydrogen (H₂), we can know the  rate of change of ammonia, that must be negative for consumption:

[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]

Nevertheless, the absolute magnitude will be positive:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Best regards.

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of Bacteria and Archaea

Answers

Answer:

the answer is monerans

Explanation:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.

What kingdom of Monera ?

Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker  and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.

Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.

Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese .  Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.

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Your question is incomplete. But your complete question is as follows:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of  _____  into  Bacteria and Archaea.

Chemistry question. Image attached.

Answers

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.

Calculate the number of moles of C2H6 in 3.97×1023 molecules of C2H6.

Answers

3.97×1023 molecules C2H6          1 mol  C2H6  

------------------------------------------ x ------------------------------------   = 0.66 mol C2H6

                                                    6.022 x 1023 molec. C2H6

Silver crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?

Answers

Answer:

Edge length of the unit cell is 4.07x10⁻¹⁰m

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

That means edge lenght is = √8 r

adius

As radius of Silver is 144pm = 144x10⁻¹²m:

a = √8 r

a = √8 ₓ 144x10⁻¹²m

a = 4.07x10⁻¹⁰m

Edge length of the unit cell is 4.07x10⁻¹⁰m

A 10.0 mL sample of calcium hydroxide solution required 26.85 mL of 0.225 M hydrochloric acid for neutralization. The balanced equation is:

Answers

Answer:

[tex]C_{base}=0.302M[/tex]

Explanation:

Hello,

In this case, we can evidence that when calcium hydroxide solution reacts with hydrochloric acid solution, the balanced neutralization reaction turns out:

[tex]2HCl(aq)+Ca(OH)_2\rightarrow CaCl_2(aq)+2H_2O(l)[/tex]

Moreover, the concentration of neutralized calcium hydroxide can be computed by using the 2:1 mole ratio between the base and the acid:

[tex]C_{acid}V_{acid}=2*C_{base}V_{base}\\\\C_{base}=\frac{C_{acid}V_{acid}}{2*V_{base}} =\frac{0.225M*26.85mL}{2*10.0mL}\\ \\C_{base}=0.302M[/tex]

Regards.

If 5.00 mL of a 0.5 M solution is diluted to a final volume of 100.0 mL, what is the concentration of the final dilute solution?

Answers

Answer:

0.025 M

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 5 mL

Initial concentration (C1) = 0.5 M

Final volume (V2) = 100 mL

Final concentration (C2) =..?

Using the dilution formula, we can obtain the final concentration of the diluted solution as follow:

C1V1 = C2V2

0.5 x 5 = C2 x 100

Divide both side by 100

C2 = (0.5 x 5)/100

C2 = 0.025 M

Therefore, the final concentration of the diluted solution is 0.025 M

The concentration of the final diluted solution is 0.025M

The dilution formula is expressed according to the formula:

[tex]C_1V_1=C_2V_2[/tex]

Given the following parameters

[tex]C_1=0.5M\\V_1=5.00mL\\V_2=100.0mL\\C_2=?[/tex]

Substitute the given parameters into the formula:

[tex]C_1V_1=C_2V_2\\0.5(5)=100C_2\\2.5=100C_2\\C_2=\frac{2.5}{100}\\C_2= 0.025M[/tex]

Hence the concentration of the final diluted solution is 0.025M

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If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu

Answers

Answer:

Copper (II) chloride.

Explanation:

Hello,

In this case, considering the described reaction which is also given as:

[tex]2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu[/tex]

For us to identify the limiting reactant we first compute the available moles of aluminium:

[tex]n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl[/tex]

Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:

[tex]n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2} =5.69molAl[/tex]

Thereby, we can infer aluminium is in excess since less moles are consumed than available whereas the copper (II) chloride is the limiting reactant.

Best regards.

What is a ‘control’ in an experiment?
A. A version of the experiment that is unchanged to make sure the experimental data is not due to chance.
B. A person who oversees the experiment to make sure it is following proper procedures.
C. The variable controlled by the scientist to affect the dependent variable.
D. The name for the set of independent and dependent variables that will be controlled by the scientist.



need help asap got 1 minute

Answers

D. The name for the set of independent and dependent variables that will be controlled by the scientist.

The statement, that describes the ‘control’ in an experiment is "the name for the set of independent and dependent variables that will be controlled by the scientist."

What is a control in experiment?

A control is an element in an experiment that remains intact or unaffected by other variables. An experiment or observation aiming to minimise the influence of variables other than the independent variable is referred to as a scientific control. It serves as a standard or point of reference against which other test findings are measured.

In a scientific experiment, an independent variable is the variable that is modified or manipulated in order to assess the effects on the dependent variable. In a scientific experiment, the dependent variable is the variable that is being tested and measured. The designation given to the set of independent and dependent variables that the scientist will regulate.

Hence the correct option is D.

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Discuss the contrary aspect of proton NMR and C-13 NMR by elucidating the structure of 2 chloro pentanal?

Answers

Answer:

See explanation

Explanation:

1H NMR

In the 2-chloro-pentanal we have 4 different types of hydrogens. Therefore, we will have 4 different signals. (See figure 1)

Red hydrogen

For the red hydrogens we have only 1 neighbor. So, if we follow the n+1 rule we can calculate the multiplicity of this hydrogen. In this case a doublet.

Blue hydrogens

In this case, we have 3 neighbors (one in the right, two in the left). Therefore we will have a quartet.

Purple hydrogens

For these hydrogens, we have also will have a quartet, because we have 3 neighbors (one in the right, two in the left).

Green hydrogens

In the green hydrogen,s we have 5 neighbors (2 in the right 3 in the left). Therefore a sextet would be produced.

Orange hydrogens

Finally, in these hydrogens, we have 2 neighbors. Therefore a triplet is expected.

13C NMR

For the 13C NMR, we have again 4 different kinds of carbons. Therefore we will have 4 signals. The most deshielded carbon, in this case, is the red one (see figure 2), so this carbon would be on the left side (around 190). Then the next deshield carbon is the blue one, due to the "Cl" atom placed on this carbon.

I hope it helps!

Read the following passage and find the two errors. Then, choose the answer that corrects the errors.
pH is a measure of the concentration of OH ions in a solution of an acid or base. The pH scale plots the concentration of solutions in a range from 0-16.
O pH is a measure of the concentration of Hions in a solution of an acid or base. The pH plots the concentration of solutions in a range from 0-14.
O pH is a measure of the concentration of H* ions in a solution of an acid or base. The basic scale plots the concentration of solutions in a range from 0-16.
O pH is a measure of the concentration of OH" ions in a solution of water. The pH scale plots the concentration of solutions in a range from 0-12
O pH is a measure of the concentration of OH" ions in a solution of an acid or base. The acid scale plots the concentration of solutions in a range from 0-
16

Answers

Answer:

pH is a measure of the concentration of H+ ions in a solution of an acid or base. The pH plots the concentration of solutions in a range from 0–14.

Explanation:

The pH is a measure of the hydrogen ion(H^+) concentration in an acid or base. It can be obtained mathematically by the formula:

pH = —Log [H^+]

The pH scale ranges from 0 to 14

Answer:

it really is A

Explanation:

just got wrong answer because i put 16 and clearly b and c makes no sence : )

The compound barium nitrate is a strong electrolyte. Write the transformation that occurs when solid barium nitrate dissolves in water.

Answers

Answer:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

Explanation:

A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.

In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.

Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.

Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.

The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.

The transformation can be represented as follows:

Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)

In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.

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The thermochemical equation is for the reaction of hydrogen bromide gas to form hydrogen gas and bromine liquid. 2HBr(g) = H 2 (g)+ Br 2 (l) 72.6 kJ How many grams of HBr (g) would be made to react if 11.4 energy were provided?

Answers

Answer:

the mass of HBr that would react is 25.41 g of HBr

Explanation:

attached is the calculations.

Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence

Answers

Considering the definition of bond and the different type of bonds, valence is not one of  the types of bonds.

What is a chemical bond

A chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.

In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.

Covalent bond

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.

Ionic bond

An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.

Metallic bond

Metallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.

Summary

In summary, valence is not one of  the types of bonds. The types of bonds are covalent, ionic and metallic.

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How many valence electrons are in the electron dot structures for the elements in group 3A(13)?

Answers

Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

What are Groups in the Periodic Table?

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.

What does the empirical formula tell you about the compound?

The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?

Can you see what type of functional group this compound could have?

Answers

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]

4. Compute the moles of oxygen by using its molar mass:

[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]

6. Search for the closest whole number (in this case multiply by 2):

[tex]C_3H_6O_2[/tex]

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

[tex]M=12*3+1*6+16*2=74g/mol[/tex]

Which is about three times in the molecular formula, for that reason, the actual formula is:

[tex]C_9H_{18}O_6[/tex]

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the energy level that the transition originated. Transitions in the Balmer series all terminate n=2.

Delta E= -2.178 x10-18J ( 1/n2Final - 1/n2Initial )

The number is 5.

What is the energy of this transition in units of kJ/mole? ( hint: the anser is NOT 4.58x10-22kJ/mole or -4.58x10-22kJ/mole)

Answers

Answer:

i. n = 5

ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])

    (1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])

⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]

But, [tex]n_{final}[/tex] = 2

434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]

434 × [tex]10^{-9}[/tex]  × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]

⇒ [tex]n_{initial}[/tex] = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])

    = (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])

    = 4.58 × [tex]10^{-19}[/tex] J

    = 4.58 × [tex]10^{-22}[/tex] KJ

But 1 mole = 6.02×[tex]10^{23}[/tex], then;

energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])

         = 7.61 × [tex]10^{-46}[/tex] KJ/mole

The initial energy level is 5  and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole

We must first calculate ΔE as follows;

ΔE = hc/λ

h = Plank's constant = 6.6 * 10^-34 Js

c = speed of light = 3 * 10^8 m/s

λ = wavelength = 434 * 10^-9

ΔE =  6.6 * 10^-34 * 3 * 10^8/434 * 10^-9

ΔE = 0.0456 * 10^-17 J

ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]

Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23

= 7.57 * 10^-43 kJ/mole

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merits of modern periodic table?​

Answers

Answer:

Merits of modern periodic table:The wrong position of some elements like argon, potassium, cobalt and nickel due to atomic weights have been solved by arranging the elements in the order of increasing atomic number without changing their own places.The isotopes of some element have the same atomic numbers. Therefore, they find the same position in periodic table.It separates metals from non-metals.The groups of the table are divided into sub groups A and B due to their dissimilar properties which make the study of elements specific and easier.The representative and transition elements have been separated.

Hope this helps...

Good luck on your assignment...

When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC

Answers

Answer:

Explanation: M(PCL5)= 31 + 5(35.5)

=208.5g/mol

M(H20)= 18g/mol

n(PCL5) = 75.5÷208.5

= 0.362mol

n(HCl)/n(PCL5)= 5/1

n(HCl)= 5×0.362

=1.81mol of HCl

Which example involves a phase change in which heat energy is released by the substance?
Ofreezing ice cream
O cooking a pot of soup
O melting ice under sunlight
O watching frost disappear into air

Answers

Answer:

Cooking a pot of soup

Explanation:

id say that because when you freeze ice cream, its already frozen, so no heat is being released. melting ice wouldn't be the answer because, once again, it is already frozen, and no heat is being released.

Answer:

the correct answer is freezing ice cream

Explanation:

i took the test & got this question correct. also, heat energy is released when freezing because there is no heat energy involved.

Calculate the mass of a body
Whose volume is
Is 2cm3 and
density is 520cm3​

Answers

Answer:

The answer is

1040g

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.

Answers

Answer:

2ErF3 + 3Mg → 2Er + 3MgF2

Explanation:

Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.

Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2

If 3.10 moles of P4010 reacted with excess water, how many grams of H3PO4
would be produced?
P4010 +6H20 + 4H3PO4
You Answered
126 g
0 0.007918
Correct Answer
O 1220 g
0.1278
75.98

Answers

Answer:

1.22 × 10³ g

Explanation:

Step 1: Write the balanced equation

P₄O₁₀ + 6 H₂O ⇒ 4 H₃PO₄

Step 2: Calculate the moles of H₃PO₄ produced by 3.10 moles of P₄O₁₀

The molar ratio of P₄O₁₀ to H₃PO₄ is 1:4. The moles of H₃PO₄ produced are 4/1 × 3.10 mol = 12.4 mol

Step 3: Calculate the mass corresponding to 12.4 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

[tex]12.4 mol \times \frac{97.99g}{mol} = 1.22 \times 10^{3} g[/tex]

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.

H2(g) → 2H+(aq) + 2e- oxidation

Cu2+(aq) → + 2e-Cu(s) reduction

Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.

Answers

Answer:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Explanation:

In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.

In the reactions:

H₂(g) → 2H⁺(aq) + 2e⁻ oxidation

Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.

The sum of both reactions is:

H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)

Subtracting the electrons in both sides of the reaction:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Answer these questions, please.

Answers

Answer:

1a. 0.89 gcm¯³

1b. Yes.

1c. Tetrahydrofuran.

2. 0.54 g/mL

Explanation:

1. Data obtained from the question include:

Volume = 0.988 L = 988 cm³

Mass = 879 g

1a. Determination of the density

Density = mass /volume

Density = 879/ 988

Density = 0.89 gcm¯³

Therefore, the density of the liquid is 0.89 gcm¯³

1b. From the given data, it is possible to determine the identity of the liquid.

1c. The density of the liquid is 0.89 gcm¯³. Comparing the density of the liquid obtained with those given in the table, the liquid is tetrahydrofuran

2. Data obtained from the question include:

Mass of empty cylinder = 5.25 g

Mass of cylinder and sodium thiosulfate = 75.82 g

Volume = 130.63 mL

Next, we shall determine the mass of sodium thiosulfate. This can be obtain as follow:

Mass of empty cylinder = 5.25 g

Mass of cylinder and sodium thiosulfate = 75.82 g

Mass of sodium thiosulfate =.?

Mass of sodium thiosulfate = Mass of cylinder and sodium thiosulfate – Mass of empty cylinder

Mass of sodium thiosulfate = 75.82 – 5.25

Mass of sodium thiosulfate = 70.57 g

Finally, we shall determine the concentration of the sodium thiosulfate as follow:

Mass = 70.57 g

Volume = 130.63 mL

Concentration =?

Concentration = mass /volume

Concentration = 70.57/130.63

Concentration = 0.54 g/mL

The concentration of the solution is 0.54 g/mL

If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?

Answers

Answer:

the density if vinegar will also be needed

Explanation:

Because this is an experiment of volumetric analysis

The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles of salt in one tablespoon.

Finally, solve (remember significant figures):

Answers

Answer:

0.109 mol/tablespoon

Explanation:

6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)

Answer:

A: 0.109

Explanation:

Edge 2020

Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1, select all the products (in formulae and general chemical terms) that could result from homolysis. In Part 2, select the reaction conditions that are most likely to promote homolysis.
Part 1. Choose all that may occur as possible products of a homolysis reaction.
Choose one or more:_______.
a. hydride ion
b. R3CO
c. Br2
d. H
e. a carbocation
f. H3C
g. H3CO-
h. hydrogen ion
i. a carbon free radical
Part 2. Choose the conditions under which homolysis is likely to occur.
Choose one or more:_______.
a. strong base
b. ultraviolet irradiation
c. high temperature
d. strong acid
e. infrared irradiation
f. low temperature

Answers

Answer:

1) R₃CO , H, H₃C, a carbon free radical

2) high temperature, ultraviolet irradiation

Explanation:

1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :

R₃CO

H

H₃C

a carbon free radical

2) Homolysis require high temperature, ultraviolet irradiation.

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