Hydrogengasand oxygengas react to form water vapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .

Answer:

Tapeworm, also called cestode, any member of the invertebrate class Cestoda (phylum Platyhelminthes), a group of parasitic flatworms containing about 5,000 species. ... Tapeworms also lack a circulatory system and an organ specialized for gas exchange.

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:

The molar mass of benzoic acid is 122.1 g/mol

The molar mass of hydrochloric acid = 36.5 g/mol

Explanation:

Benzoyl chloride is an organic compound with the molecular formula C₆H₅COCl. It is an acyl chloride since is it an organic derivative of a carboxylic acid. Acyl chlorides have the general molecular formula, R-COCl, where R is a side chain.

The R group of benzoyl chloride is the benzyl group C₆H₅. It reacts with water (hydrolysis) to produce hydrochloric acid and benzoic acid. The equation of the reaction is given below:

C₆H₅COCl + H₂O → C₆H₅CO₂H + HCl

The molar mass of benzoic acid as well as of hydrochloric acid is calculated from the sum of the masses of the atoms of the elements present in the compound thus:

Molar mass of carbon = 12.0107 g

Molar mass of hydrogen = 1.00784 g

Molar mass of oxygen = 15.999 g

Molar mass of chlorine = 35.453 g

Molar mass of benzoic acid, C₆H₅CO₂H containing 7 moles of atoms of carbon, 6 moles of atoms of hydrogen and 2 moles of atoms of oxygen = 7 × 12.0107 + 6 × 1.00784 + 2 × 15.999 = 122.1 g

Therefore, the molar mass of benzoic acid is 122.1 g/mol

Molar mass of hydrochloric acid, HCl, containing 1 mole of atoms of hydrogen and 1 mole of atoms of chlorine = 1 × 1.00784 + 1 × 35.453 = 36.5 g

Therefore, the molar mass of hydrochloric acid = 36.5 g/mol

Answer:

Systems move from ordered behavior to more random behavior.

Explanation:

Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)

According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.

Answer:

The lewis structure for NO₃⁻ is shown in the attachment below

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Explanation:

The lewis structure for NO₃⁻ is shown in the attachment below.

From the Lewis structure

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

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:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

Answer:

d

Explanation:

since it is much convenient since the email will not get lost and it's contents will not be forgotten

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer:

250ml

Explanation:

call it V

V*0.2=0.05 (moles)

so V=0.05/0.2 = 0.25l = 250ml

We know

[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=\dfrac{0.05}{0.2}[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=0.25L[/tex]

[tex]\\ \Large\sf\longmapsto Volume\:of\:KI=250mL[/tex]

Answer: The mass of copper (II) nitrate produced is 105.04 g.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of copper = 35.5 g

Molar mass of copper = 63.5 g/mol

Plugging values in equation 1:

[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]

The given chemical equation follows:

[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]

By the stoichiometry of the reaction:

If 3 moles of copper produces 3 moles of copper (II) nitrate

So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate

Molar mass of copper (II) nitrate = 187.56 g/mol

Plugging values in equation 1:

[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]

Hence, the mass of copper (II) nitrate produced is 105.04 g.

Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

Answer:

Mass gallium (Ga°(s)) produced ≅ 0.800 grams (1 sig. fig.)

Explanation:

Ga(OH)₃ => Ga⁺³ + 3OH⁻

Ga⁺³ + 3e⁻ => Ga°(s)

? grams Ga°(s) = 0.680 Amps x 1 mole e⁻/1 Faraday x 1 Faraday/96,500 Amp·sec x 1 mole Ga°/3 moles e⁻ x 69.723 grams Ga°/mole Ga° x 60 sec/1 min x 80 min = [(0.680)(69.723)(60)(80)/(96,500)(3)] grams Ga° = 0.786099731 grams Ga° (calc. ans.) ≅ 0.800 grams Ga°  (1 sig. fig.)

Explanation:

Ethanol is made by the hydration of ethylene in the presence of a catalyst such as sulfuric acid (H 2SO 4).

Answer:

a) N2

b) They all have the same average kinetic energy.

Explanation:

At a given temperature, the speed of a gas molecule depends on its relative molecular mass. The heavier the gas, the lesser its average velocity at a given temperature. On that basis, N2 molecules are the fastest moving gas molecules.

At a particular temperature, all gases have the same average kinetic energy.

Answer:

Blue or Purple color filter

Explanation:

Given that Sodium has an emission spectrum with wavelength ≈ 590nm and for a wavelength of 590nm  the color is yellow.

Hence To filter out the color ( yellow ) to enable the measurement of the Balmer series of hydrogen spectrum, we have to use a filter that possess the complementary color of yellow ( i.e. purple(RYB color model) or blue (RGB additive color model )

therefore color filter to be used = Blue or Purple

Answer:

There is 13.48 grams of P2O5 formed

Explanation:

Step 1: Data given

A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.

Mass of P = 5.89 grams

Molar mass of O2 = 32.0 g/mol

atomic mass of P = 30.97 g/mol

molar mass of P2O5 = 141.94 g/mol

Step 2: The balanced equation

4P(s)+5O2(g)⇔ 2P2O5(s)

Step 3: Calculate moles of P

Moles P = Mass P / atomic mass P

Moles P = 5.89 grams / 30.97 g/mol

Moles P = 0.190 moles

Step 4: Calculate moles of P2O5

For 4 moles P we need 5 moles O2 to produce 2 moles P2O5

For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5

Step 5: Calculate mass of P2O5

Mass P2O5 = moles P2O5 * molar mass P2O5

Mass P2O5 = 0.095 moles * 141.94 g/mol

Mass P2O5 = 13.48 grams

There is 13.48 grams of P2O5 formed

Answer:

The final volume of the sample of gas is 36.287 liters.

Explanation:

Let suppose that sample of gas is a closed system, that is, a system with no mass interactions with surroundings, and gas is represented by the equation of state for ideal gases, which is described below:

[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)

Where:

[tex]P[/tex] - Pressure, in atmospheres.

[tex]V[/tex] - Volume, in liters.

[tex]n[/tex] - Molar quantity, in moles.

[tex]T[/tex] - Temperature, in Kelvin.

[tex]R_{u}[/tex] - Ideal gas constant, in atmosphere-liters per mole-Kelvin.

As we know that sample of gas experiments an isobaric process, we can determine the final volume by the following relationship:

[tex]\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}[/tex] (2)

Where:

[tex]V_{1}[/tex] - Initial volume, in liters.

[tex]V_{2}[/tex] - Final volume, in liters.

[tex]T_{1}[/tex] - Initial temperature, in Kelvin.

[tex]T_{2}[/tex] - Final temperature, in Kelvin.

If we know that [tex]V_{1} = 33\,L[/tex], [tex]T_{1} = 291.15\,K[/tex] and [tex]T_{2} = 320.15\,K[/tex], then the final volume of the gas is:

[tex]V_{2} = V_{1}\cdot \left(\frac{T_{2}}{T_{1}} \right)[/tex]

[tex]V_{2} = 33\,L \times \frac{320.15\,K}{291.15\,K}[/tex]

[tex]V_{2} = 36.287\,L[/tex]

The final volume of the sample of gas is 36.287 liters.

Answer:

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Answer:

answer is A. Endocrine system

Endocrine glands secrete hormones straight into the bloodstream. Hormones help to control many body functions, such as growth, repair and reproduction.

Answer:

A endocrine system

this is the answer

Explanation:

Following are the kinetic theory of gases postulates:

1) Space-volume to molecules ratio is negligible.

2)There is no force of attraction between the molecules at normal temperature and pressure. The force of attraction between the molecules build when the temperature decreases and the pressure increases.

3) There is large space between the molecules resulting in continuous motion.

4) The free movement of molecules results in collision which is perfectly elastic.

5) The molecules have kinetic energy due to random movement. But the average kinetic energy of these molecules differs with temperature.

6) Molecules exert pressure on the walls of the container.

Answer:

the answer is C

Explanation:

a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule

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Answer:

The number of moles of strontium sulfide produced is:

= 1.663.

Explanation:

Chemical formula for strontium sulfide = SrS

Production of strontium sulfide = 199g

1 mole = 1 moles Strontium Sulfide, which is equal to 119.685 grams

The number of moles of strontium sulfide produced = 1.663 (199/119.685)

The number of moles of strontium sulfide produced is the dividend of the amount of strontium sulfide produced during the experiment divided by the mass of 1 mole.

Answer:

5g/cm3

Explanation:

firstly convert the litres and kilograms to grams and centimeters.

1l is equivalent to 1000cm3

0.58×1000

580cm3

and 1kg is equivalent to 1000g

2.9×1000

2900

then find the density by using the formula

density=mass/volume

=2900g/580cm3

=5g/cm3

I hope this helps

Answer:

Average specific heat capacity of metal = 0.57 J/g°C

Explanation:

Heat lost = Heat gained

Heat energy gained or lost, H = mcΔT

where m = mass of substance, c = specific heat capacity, ΔT = temperature change

Trial 1:

Heat lost by metal = -[2.746 g × c × ΔT]

ΔT = (26.3 - 72.1) °C = -45.8 °C

Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C

Heat gained by water = 15.200 × 4.18 × ΔT

ΔT = (26.3 - 24.7) = 1.6 °C

Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J

From Heat lost = Heat gained

c × (125.7688)g°C = 101.6576 J

c = 101.6576 J / 125.7688 g°C

c = 0.8083 J/g°C

Trial 2:

Heat lost by metal = -[2.750 g × c × ΔT]

ΔT = (26.2 - 72.2)°C] = - 46 °C

Heat lost by metal = -[2.750 g × c × (-46 °C)

Heat lost by metal = c × (126.5) g°C

Heat gained by water = 15.206 × 4.18 × ΔT

ΔT = (26.2 - 24.6) = 1.6 °C

Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J

From Heat lost = Heat gained

c × (126.5)g°C = 101.6977 J

c = 101.697728 J / 126.5 g°C

c = 0.8039 J/g°C

Trial 3:

Heat lost by metal = -[2.900 g × c × ΔT]

ΔT = (24.7 - 71.9)°C] = - 47.2 °C

Heat lost by metal = -[2.900 g × c × (- 47.2 °C)

Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C

Heat gained by water = 15.201 × 4.18 × ΔT

ΔT = (24.7 - 24.5) = 0.2 °C

Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J

From Heat lost = Heat gained

c × (136.88)g°C = 12.708036 J

c = 12.708036 J / 136.88 g°C

c = 0.0928 J/g°C

Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3

Average specific heat capacity of metal = 0.57 J/g°C

Explanation:

Because molarity is mol/L, we'll have to convert 17g to mol.

After obtaining the mol, we'll divide that by the volume to obtain Molarity.

[tex]\\ \sf\longmapsto NH_3[/tex]

[tex]\\ \sf\longmapsto 14u+3(1u)[/tex]

[tex]\\ \sf\longmapsto 14u+3u[/tex]

[tex]\\ \sf\longmapsto 17u[/tex]

[tex]\\ \sf\longmapsto 17g/mol[/tex]

Moles of Ammonia:-

[tex]\boxed{\sf No\:of\:moles =\dfrac{Given\:mass}{Molar\:mass}}[/tex]

[tex]\\ \sf\longmapsto No\:of\:moles=\dfrac{17}{17}[/tex]

[tex]\\ \sf\longmapsto No\;of\:moles=1mol[/tex]

We know

[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\: Solution\:in\:L}}[/tex]

[tex]\\ \sf\longmapsto Molarity=\dfrac{1}{0.50}[/tex]

[tex]\\ \sf\longmapsto Molarity=2M[/tex]

Answer:

Robert Millikan was a physicist who discovered the elementary charge of an electron using the oil-drop experiment

Answer:

the mass of an electron using the Oil-Drop experiment.

Explanation:

Answer:

Explanation:

nFeSo4=3.36/152

nkmno4=1/5nFeSO4

V=17.68 ml