5.96 g of ammonia reacts completely according to the following reaction:
2 NH, (g) + Co, (g) → CN,OH, (s) + H20 (1)
(a) What is the theoretical yield of urea (CN,OH,) for this reaction?
(b) If 13.74 g of urea are produced, what is the percent yield for this equation?

please show work, will give brainliest

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Which must be balanced according to the law of conservation of mass:

[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]

Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.

Regards!

Answer:

See explanation

Explanation:

The molecule called guanidine is shown in the image attached to this answer. It contains three nitrogen atoms. Two among them are sp3 hybridized while one of them is sp2 hybridized.

Guanidine is more basic than other amines because its protonanation leads to the formation of three equivalent resonance structures thereby making its protonated form quite stable. This effect is not observed in other amines.

Also, the sp2 hybridized nitrogen atom is more basic and more easily protonated because when it is protonated, three equivalent resonance structures are obtained.

Answer:

the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer

Answer:

See explanation

Explanation:

Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.

Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.

The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.

Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.

The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.

The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

The molal concentration of the resulting solution is 0.492 m. The concentration number of moles of solute in one kg of solvent.

It is the measure of the concentration number of moles of solute in one kg of solvent.

To calculate the molal concentration first calculated the mass of [tex]\bold {LiCl}[/tex],

[tex]\rm \ mass = density \times volume[/tex]

Put the values,

m = 1 g/mL x  19.7 mL

m = 19.7 g     or

m = 0.0197 kg

Calculate the moles of LiCl:

[tex]n =\rm \dfrac { 0.411 \ g \times 1 mol }{ 42.394 \ g }\\\\n = 9.69\times 10^{-3} \ moles[/tex]

So, now the molal concentration,

[tex]m = \rm \dfrac { 9.69\times 10^{-3}\ mol} {0.0197 \ kg} \\\\m = 0.492[/tex]

Therefore, the molal concentration of the resulting solution is 0.492 m.

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Answer:

When the negatively-charged rod is brought close to the electroscope, positive charges are attracted to it and negative charges are repelled away from it. The electroscope has a net neutral charge and the rubber rod has a net negative charge. If they are brought into contact, they will both take a net negative charge.

Explanation:

I looked it up

Answer:

46 pounds

Explanation:

Let the amount of Plutonium-238 left after 10 years be P

Let the initial mass of Plutonium-238 be Po

Let the rate of decay be r

Let the time taken be t

Hence;

P = Poe^-rt

P= 50e^-0.008×10

P= 46 pounds

Answer:

h2o2 and liver produce oxygen gas

Answer:

C

Explanation:

u can't touch a chemical with bare skin

Answer:

touching a chemical with his or her bare skin.

Answer:

96 grams

Explanation:

Answer:

e) C5H10O3

Explanation:

According to the information given in this question;

50.84% represents C

8.53% represents H

[100% - (50.84 + 8.53)]

= 100 - 59.37

= 40.63% of Oxygen (O)

This percentage means that;

Carbon = 50.84g

Hydrogen = 8.53g

Oxygen = 40.63g

By dividing by their respective atomic masses, we convert to moles:

C = 50.84g ÷ 12 = 4.24mol

H = 8.53g ÷ 1 = 8.53mol

O = 40.63g ÷ 16 = 2.54mol

Next we divide by the smallest no of the values (2.54mol)

C = 4.24 ÷ 2.54 = 1.669

H = 8.53 ÷ 2.54 = 3.358

O = 2.54 ÷ 2.54 = 1

To get a simple whole number ratio, we multiply the results by 3

C = 1.669 × 3 = 5.007

H = 3.358 × 3 = 10.074

O = 1 × 3 = 3

Simple whole number ratio of carbon, hydrogen and oxygen is 5:10:3. Hence, the empirical formula is C5H10O3.

Answer: photosynthesis

Explanation:

Answer:

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

[tex]V_{gas}=\sqrt{\frac{8RT}{\pi M}}[/tex]

     OR

[tex]V_{gas}\propto \sqrt{\frac{1}{M}}[/tex]

where, M is the molar mass of gas

Forming an equation for the two gases:

[tex]\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}[/tex]          .....(1)

Given values:

[tex]V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol[/tex]

Plugging values in equation 1:

[tex]\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s[/tex]

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

[tex]Rate\propto \frac{1}{\sqrt{M}}[/tex]

Where, M is the molar mass of the gas

Forming an equation for the two gases:

[tex]\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}[/tex]          .....(2)

Given values:

[tex]Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol[/tex]

Plugging values in equation 2:

[tex]\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr[/tex]

Hence, the rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

A colloid is a mixture in which one substance of microscopically dispersed insoluble particles are suspended throughout another substance. However, some definitions specify that the particles must be dispersed in a liquid, and others extend the definition to include substances like aerosols and gels.

The types of colloids includes sol, emulsion, foam, and aerosol.

Explanation:

everything can be found in the picture

Answer:

turtle

Explanation:

they are slow and they take there time

Answer:

The pressure will be 933.33 Kpa

Explanation:

Given that:

Volume V₁ = 200 cm³  (note, there is a mistake in the volume. It is supposed to be 200 cm³)

Pressure P₁ = 700 Kpa

Pressure P₂ = ??? (unknown)

Volume V₂ = 150 cm³

Temperature = constant

Using Boyle's law:

PV = constant

i.e.

P₁V₁ = P₂V₂

700 Kpa × 200 cm³ = P₂ × 150 cm³

P₂ = (700 Kpa × 200 cm³)/150 cm³

P₂ = 933.33 Kpa

Answer:

Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.

Note: The volume of NaOH used isnot given and issp to be determined by titration.

Explanation:

The question is a lab activity. Therefore, the volume of 0.2 M NaOH required for the neutralization of 30.0 mL of hydrochloric acid of unknown concentration is to be determined from titration. However, assuming a certain volume of the base is used for the complete neutralization of the acid, the concentration of the acid can be calculated.

Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid can be calculated thus:

Step 1: Balanced chemical equation of reaction:

HCl + NaOH ---> NaCl + H₂O

From the equation of the reaction, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and water.

Step 2: Determine the number of moles of NaOH in 20.0 mL of 0.200 M solution

Molarity = number of moles/volume in Litres

Number of moles = molarity × volume in litres

Volume of NaOH in litres = 20 ml × 1 litre/1000ml = 0.02 litres

Number of moles NaOH = 0.02 L × 0.2 M = 0.004 moles

Therefore, 0.00 M of NaOH reacted with 0.004 moles of HCl

Step 3: Determine concentration of HCl

Molarity or concentration = number of moles / volume in litres

Number of moles of HCl = 0.004 moles

Volume of HCl in litres = 30.0 mL × 1 L/1000 mL = 0.03 L

Concentration of HCl = 0.004 moles / 0.03 L = 0.133 M

Therefore, assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.

Answer:

Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

Explanation:

Oxidation:                            Al°(s) =>   Al⁺³(aq) + 3e⁻

Reduction:           3Ag⁺(aq) + 3e⁻ => 3Ag°(s)

_________________________________________

Net Rxn:           Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

One mole of neutral aluminum atoms (Al°(s)) undergo oxidation delivering 3 moles  of electrons to 3 moles silver ions (3Ag⁺³(aq)) that are reduced to 3 moles of neutral silver atoms (3Ag°(s)) in basic standard state 25°C; 1atm.

A balanced equation obeys the law of conservation of mass. According to the law of conservation of mass, mass can neither be created nor be destroyed. The coefficient of silver is 3.

A balanced chemical equation can be defined as the chemical equation in which the number of reactants and products on both sides of the equation are equal. The amount of reactants and products on both sides of the equation will be equal in a balanced chemical equation.

The numbers which are used to balance the chemical equation are called the coefficients. The coefficients are the numbers which are added in front of the formula.

The balanced chemical equation for the given redox reaction is given as:

Al (s) + 3 Ag⁺ (aq)  → Al³⁺ (aq) + 3Ag (s)

Thus the coefficient of silver is 3.

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Answer:

Federal law is a law that may protect people from digital attacks such as cyberbullying.

Explanation:

Although it is mostly based on the situation, federal law does protect those who suffer from cyberbullying. The attackers may be charged with defamation, since cyberbullying consists of spreading false information about someone on the internet, that's when the law comes in.

Answer:

(a) The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]

(b)The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]

Explanation:

The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

Molality is expressed in units [tex]\frac{moles}{kg}[/tex]

(a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 [tex]\frac{g}{mole}[/tex], then 14.3 grams of the compound represents the following number of moles:

[tex]14.3 grams*\frac{1 mole}{342 grams} =[/tex] 0.042 moles

Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:

[tex]molality=\frac{0.042 moles}{0.685 kg}[/tex]

Solving:

molality= 0.0613[tex]\frac{moles}{kg}[/tex]

The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]

(b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:

[tex]molality=\frac{7.15 moles}{3.505 kg}[/tex]

Solving:

molality= 2.04[tex]\frac{moles}{kg}[/tex]

The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]

Answer:

the Molecular formula will be; C10H14O

Explanation:

Given the data in the question;

Chrysanthenone is an unsaturated ketone,

it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).

molecular formula = ?

we know that ketone contain 1 oxygen and mass of oxygen is 16

so mass of the C and H remaining will be;

⇒ 150 - 16 = 134

Now we determine the number of C atoms;

⇒ 134 / 13 = 10

hydrocarbon with 10 hydrogen atom have CnH2n+2 means

⇒ ( 10 × 2 ) +2 = 22 hydrogens

But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens

⇒ 22 - 6 - 2 = 14

Hence the Molecular formula will be; C10H14O

Explanation:

Do đó cấu hình electron của canxi là: 1s2 2s2 2p6 3s2 3p6 4s2.

Answer:

2266g

Explanation:

mass = no.of molecules /6.o23*1o(23) * molar mass

molar mass of co2= 44g /mol

1.5 .10^25/6.023 .10^23 =51.5 moles of co2

51.5 .44g/mol =2266 g

1 mole is equal to 6.023 × 10 ²³ molecules. 2266 grams are in 1.50 x 10²⁴ atoms of copper.

The term mole is defined as the amount of substance of a system which contains as many elementary entities.

One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.

The mole concept can be used to convert between mass and number of particles.

mass = number of molecules / 6.023 × 10²³ × molar mass

molar mass of CO2 = 44g /mol

1.5 .10^25 / 6.023 × 10^23

= 51.5 moles of CO2

= 51.5 .44g/mol

= 2266 g

Thus, 2266 grams are in 1.50 x 10²⁴ atoms of copper.

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Answer:

40 grams of KCI are required to make 500 ml of 8% solution.

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

For the given reaction, 2 moles of C₆H₆ the heat energy released is - 6535 KJ. Then, for 16 g of the compound or 0.205 moles needs 669.83 KJ of heat released in combustion.

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, typically oxygen, resulting in the release of heat, light, and various combustion products, such as carbon dioxide and water vapor.

The process of combustion involves a rapid and exothermic (heat-releasing) oxidation reaction that produces a flame, which is visible in many cases.

Here, 2 moles of the hydrocarbon releases - 6535 KJ of energy.

molar mass of C₆H₆ = 78 g/mol

then no.of moles in 16 g = 16 /78 = 0.205 moles.

then energy released by 0.205 moles  = 0.205 moles × 6535 KJ /2 moles = 669.83 kJ

Therefore, the heat energy released by 16 g of the compound in combustion is 669.83 kJ.

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Answer:

verdadero

Explanation:

porque esoo [tex]\lim_{n \to \infty} a_n x_{123} \frac{x}{y} \sqrt[n]{x} x^{2} \sqrt{x} \pi \neq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right.[/tex] 

Answer: hello your question is poorly written below is the complete question

answer:

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Explanation:

Determining the type of orbital in which the lone pair on each N atom will reside.

From the configuration attached below we can determine the type of orbital and they are ;

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital