Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with ammonia NH3. What mass of ammonium phosphate is produced by the reaction of 5.5g of phosphoric acid

Answer:

23

Explanation:

The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.

The equation below gives the number of atoms present at time t

[tex]N=Noe^-kt[/tex]

N = Number of atoms present at time t

No = Number of atoms initially present

k = decay constant

t = time taken

Given that;

t1/2 = 0.693/k

where t1/2 = half life

k = 0.693/t1/2

k = 0.693/ 55.6 s

k = 0.0125 s-1

Substituting values;

N = 16,000 e^-0.0125(0)

N = 16,000 atoms

At 50 s

N = 16,000 e^-0.0125(50)

= 8564 atoms

At 100 s

N = 16,000 e^-0.0125(100)

= 4584 atoms

At 200 s

N = 16,000 e^-0.0125(200)

= 1313 atoms

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Answer:

The answer is A. +1

Explanation:

Answer:

p'PCl3 =  6.8 torr

p'Cl2 =26.4 torr

p'PCl5 =223.4 torr

Explanation:

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is

PCl3(g) + Cl2(g) ---> PCl5(g)

Calculate the new partial pressures after equilibrium is reestablished. [in torr]

pPCl3

pCl2

pPCl5

Step 1: Data given

Partial pressure before adding chlorine gas:

Partial pressure of PCl5 = 217.0 torr

Partial pressureof PCl3 = 13.2 torr

Partial pressureof Cl2 = 13.2 torr

A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing

Step 2: The equation

PCl3(g)+Cl2(g) ⇔ PCl5(g)

Step 3: The expression of an equilibrium constant before adding chlorine gas

Kp = pPCl5 / (pPCl3 * pCl2)

Kp = 217.0 / (13.2 * 13.2)

Kp =  1.245

Step 4:  The expression of an equilibrium constant after adding chlorine gas

Partial pressure of PCl5 = 217.0 torr

Partial pressure of PCl3 = 13.2

Partial pressure of Cl2 = TO BE DETERMINED

Step 5: The total pressure of the system

Ptotal = pPCl5 + pPCl3 + pCl2

263.0 torr = 217.0 torr + 13.2 torr + pCl2

pCl2 = 263.0 - 217.0 -13.2 = 32.8 torr

Step 6: The initial pressure

The equation: PCl3(g)+Cl2(g) ⇔ PCl5(g)

pPCl3 = 13.2 torr

pCl2 = 32.8 torr

pPCl5 = 217.0 torr

Step 7: The pressure at the equilibrium

p'PCl3 = (13.2 -x) torr

p'Cl2 = (32.8 - x) torr

p'PCl5 = (217.0 + x) torr

Step 8: The equilibrium constant

'Kp =  p'PCl5 / (p'PCl3 * p'Cl2)

1.245 = (217.0+x) / ((13.2-x)(32.8-x)

x = 6.40 torr

p'PCl3 = 13.2 -6.40 = 6.8 torr

p'Cl2 = 32.8 - 6.40 =26.4 torr

p'PCl5 = 217.0 + x) 6.4 = 223.4 torr

Answer:

6CO2+6H2O+heat" C6H12O6+6O2

the best way to balance a chemical reaction is to start with balancing the hydrogen followed by the other elements then lastly oxygen.so in this case if you put a 6 in front of carbon dioxide,water and oxygen you will definitely balance it.cause at the first side you have 6 carbons similar to the product,12 oxygen similar to the product and 18 oxygen similar to the products.

I hope this helps

Answer:

Explanation:

I saw this after answering your other question on the same reaction.

To balance the chemical reaction, look at the reactants and products. As O is part of both products, focus on C and H instead.

On the products side, 1 C6H12O6 has 6 C and 12 H. So that requires the same numbers of C and H on the reactant side because of mass conservation.

That gives 6 CO2 and 6 H2O as the reactants. Counting the number of O in the reactants, there are 6*2 + 6 = 18 O. Subtracting the 6 O in C6H12O6, that leaves 12 O so there are 12/2 = 6 O2 in the products.

Combining the numbers above, the balanced equation is:

___6___ CO₂ + ___6___ H₂O + heat ↔ ___1___ C₆H₁₂O₆ + ___6___ O₂

Answer:

[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]

Explanation:

We are asked to find how many grams of sodium chloride are in a solution.

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]

We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.

There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.

[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]

Now we know the molarity and the liters of solution, but the moles of solute are unknown.

Substitute the values into the formula.

[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]

We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]

The units of liters cancel.

[tex]0.250 * 0.75 \ mol \ NaCl[/tex]

[tex]\bold {0.1875 \ mol \ NaCl}[/tex]

Now that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.

Set up a ratio so we can convert using dimensional analysis.

[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

Multiply by the number of moles we calculated.

[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

The units of moles of sodium chloride cancel.

[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]

[tex]\bold {10.9575 \ g \ NaCl}[/tex]

The original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.

[tex]11 \ g \ NaCl[/tex]

There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.

Answer:

The watch is 40.9 years old.

Explanation:

To know how many years old is the watch we need to use the following equation:

[tex] I_{(t)} = I_{0}e^{-\lambda t} [/tex]   (1)

Where:

[tex]I_{(t)}[/tex]: is the brightness in a time t = (1/10)I₀

[tex]I_{0}[/tex]: is the initial brightness

λ: is the decay constant of tritium

The decay constant is given by:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]   (2)

Where:

[tex]t_{1/2}[/tex]: is the half-life of tritium = 12.3 years

By entering equation (2) into (1)  we have:

[tex] I_{(t)} = I_{0}e^{-\lambda t} = I_{0}e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]

[tex] \frac{I_{(t)}}{I_{0}} = e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]

By solving the above equation for "t" we have:

[tex] ln(\frac{I_{(t)}}{I_{0}}) = -\frac{ln(2)}{t_{1/2}}t [/tex]

[tex] t = -\frac{ln(\frac{I_{(t)}}{I_{0}})}{\frac{ln(2)}{t_{1/2}}} = -\frac{ln(\frac{1}{10})}{\frac{ln(2)}{12.3}} = 40.9 y [/tex]

Therefore, the watch is 40.9 years old.

I hope it helps you!

Answer:

1.94 × 10⁻³

Explanation:

Step 1: Calculate the concentration of H⁺ ions

We will use the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M

Step 2: Calculate the acid dissociation constant (Ka) of the acid

For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.

Ka = [H⁺]²/Ca

Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³

Answer:

acetone, (CH3)2CO cyclohexane are the compound that possesses a permanent dipole

Explanation:

Permanent dipole describes the partial charge separation that can occur within a molecule along with the bond dat form between 2 different atoms

Answer:

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

Explanation:

Step 1: Data given

The oxidation number of manganese ion (Mn2+ ) is +2

The oxidation number of manganese dioxide  is +(MnO2)4

This means the oxidation number from Mn will go from +2 to +4, since it's increased, this is an oxidation reaction

Mn2+(aq)  ⇒ MnO2(s)

We have to balance both sides. Mn is already the same. But on the right side we have O atoms. T obalance both sides we have to add O atoms to the left side. This by adding 2x H2O

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s)

Now the amount of O atoms is balanced, but we have H- atoms at the left side. To balance we have to add 4 H atoms to the right side

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq)

Now the amount of atoms is balanced at both sides. We also have to check if the charge on both sides is the same.

Since the left side has a charge of +2, and right has a charge of +4, we have to add 2 electrons to balance this.

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

Answer:

The products will be favored at equilibrium.

Explanation:

The balanced chemical equation for the reaction is the following:

2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)

The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).

The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:

[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]

Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²

So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).

Therefore, products will be favored at equilibrium at 298 K.

Answer: The thermal energy that would be released by 276.0g of sodium acetate trihydrate is 71.8kJ.

Explanation:

Supercooling is the process of lowering the temperature a liquid below its freezing point, without it becoming solid. A liquid below its freezing point will crystallize in the presence of a seed crystal because it serves as a structure for formation of crystals. From the question,

The given mass of sodium acetate trihydrate

(CH3COONa.3H2O)= 276.0g

Molar mass of sodium acetate

trihydrate= 136.08g/mol

Thermal heat of fusion of sodium acetate

trihydrate = 35.9 kJ/mol

From the given mass the number of moles present= 276.0/ 136.08

= 2.0moles

Therefore the heat (thermal) energy of the given mass of sodium acetate

trihydrate = 2.0 × 35.9

= 71.8kJ

Therefore, upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes.

Answer:

low-spin complexes contain the maximum number of unpaired electrons.

Explanation:

In the crystal field theory, the magnitude of crystal field splitting and the pairing energy determines whether a complex will be low spin or high spin.

Low spin complexes often have greater magnitude of crystal field splitting energy than low spin complexes.

High spin complexes have maximum number of unpaired electrons(most of the electrons are unpaired) while low spin complexes have a minimum number of unpaired electrons in a complex(most of the electrons are paired).

Explanation:

here's the answer to your question

Answer:

50.0 moles C02

Explanation:

First write down the CORRECTLY balanced equation. NOTE: The equation you provide is incorrect.

2C2H2(g) + 5O2(g) ==> 4CO2(g) + 2H2O(g)  CORRECT EQUATION

Next, look at the stoichiometric ratio of C2H2 to CO2.  You can see it is 2 moles C2H2 produces 4 moles CO2.

Thus, 25.0 moles C2H2 x 4 moles CO2/2 moles C2H4 = 50.0 moles CO2

Answer:

C) NH3 > PH3 > CH4

Explanation:

The boiling point of a substance depends on the nature of intermolecular interaction between the molecules of the substance. The greater the magnitude of intermolecular interaction between the molecules of the substance, the higher the boiling point of the substance.

Both NH3 and PH3 have intermolecular hydrogen bonding between their molecules. However, since nitrogen is more electronegative than phosphorus, the magnitude of intermolecular hydrogen bonding in NH3 is greater than in PH3 hence NH3 has a higher boiling point than PH3.

CH4 molecules only have weak dispersion forces between them hence they exhibit the lowest boiling point.

Answer:

b. radicals form easily in the presence of chlorine radicals.

Explanation:

Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."

This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.

Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."

Answer:

CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)

Explanation:

Hydration is the process by which anhydrous CuSO4 acquires molecules of water of crystalization to form the pentahydrate.

The water of crystalization becomes attached go the crystals of the CuSO4 to form the hydrated salt.

Beginning with solid anhydrous CuSO4 we have;

CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)

Answer:

[tex]error = 4.0 - 3.5 = 0.5 \\ \\ percent \: error = \frac{0.5}{3.5} \times 100 \\ \\ = 14.29\% [/tex]

Answer:

a. Na(s); b. Al(s); c. Zn(s)

Explanation:

Let's consider the following redox reactions.

a. ZnCl₂ (aq) + 2 Na(s) → Zn(s) + 2 NaCl(aq)

Na is oxidized because its oxidation number increases from 0 to +1 (in NaCl) whereas Zn is reduced because its oxidation number decreases from 2+ (in ZnCl₂) to 0.

b. Al(s) + FeBr₃ (aq) → AlBr₃ (aq) + Fe(s)

Al is oxidized because its oxidation number increases from 0 to +3 (in AlBr₃) whereas Fe is reduced because its oxidation number decreases from 3+ (in FeBr₃) to 0.

c. FeSO₄ (aq) + Zn(s) → Fe(s) + ZnSO₄(aq)

Zn is oxidized because its oxidation number increases from 0 to +2 (in ZnSO₄) whereas Fe is reduced because its oxidation number decreases from 2+ (in FeSO₄) to 0.

Answer:

C-O: polar covalent

Mg-F: ionic

Cl-Cl: nonpolar covalent

Explanation:

Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.

Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.

In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.

In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.

Answer:

C

Explanation:

because it remains the same

Solution :

Given :

Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm

Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm

Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm

Reaction :

[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]

Using equilibrium concept,

[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]

[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]

[tex]$k_p=0.594$[/tex]

[tex]k_p=0.59[/tex]  (in 2 significant figures)

or [tex]k_p=5.9 \times 10^{-1}[/tex]

Answer:

a. 181.5 kcal

Explanation:

Step 1: Calculate the enthalpy of the process (ΔH°).

Let's consider the following process.

CO₂(s) → CO₂(g)

We can calculate the enthalpy of the process using the following expression.

ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)

ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))

ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ

According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.

Step 2: Convert 986 g of CO₂ to moles

The molar mass of CO₂ is 44.01 g/mol.

986 g × 1 mol/44.01 g = 22.4 mol

Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂

22.4 mol × 33.9 kJ/1 mol = 759 kJ

We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.

759 kJ × 1 kcal/4.184 kJ = 181.5 kcal

Answer: Options related to your question is missing below are the missing options

a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.

b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.

c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes

d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.

e. two molecules of cytochrome c must first combine physically before they are catalytically active.

answer:

cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )

Explanation:

The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )

An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.

Answer:

The answer is "152 pm".

Explanation:

The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.

The atomic radius of [tex]H= 37.0 \ pm[/tex]

The atomic radius of [tex]Br = 115.0 \ pm[/tex]

[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]

                    [tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]

Answer:

Haematology is the specialty important for the diagnosis and management of a wide range of benign and malignant disorders of the red and white blood cells, platelets and the coagulation system in adults and children.

Answer:

groups are based on how many electrons to become stable

Explanation: