Butanal, an aldehyde, can be made from 1-bromobutane, but it requires two reactions in sequence. What are the two reagents that can be used to make butanal from 1-bromobutane if reagent A is first reacted with 1-bromobutane and then the product of that reaction, is reacted with reagent B.

I think the answer most be d

In chemistry like dissolves like hence hexane will dissolve in gasoline.

Dissolution stems from intermolecular interaction between solute and solvent molecules.

If this interaction can not occur, dissolution of one substance in another is impossible.

Hexane dissolves in gasoline because the both substances are non-polar and can interact with each other effectively.

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Answer:

C . Kinetic Molecular Theory

The correct option is d.: 1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

[tex]moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}[/tex]

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

[tex]moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}[/tex]

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

[tex]mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}[/tex]

mass of glucose= 1.5 grams

In summary, the correct option is d.:  1.5 grams of glucose is produced from 2.20 g of CO₂.

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Answer:

A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.

Examples - Superconductors and superfluids are the two examples of BEC.

Explanation:

Answer:

an if/then statement

Explanation:

Answer:

11.76 g/cm^3

Explanation:

Mass of empty flask and stopper = 64.232g

Mass of flask filled with water = 153.617 g

Mass of water = 153.617 g - 64.232g = 89.385 g

Mass of flask, stopper and metal = 143.557 g

Mass of metal = 143.557 g - 64.232g = 79.325 g

Mass of water, flask, stopper and metal = 226.196 g

Mass of water = 226.196 g - 143.557 g = 82.639 g

Since mass of water =volume of water

Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3

Density of metal = mass/volume = 79.325 g/6.746 cm^3

= 11.76 g/cm^3

Explanation:

The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.

The required volume of [tex]HNO_3[/tex] is V1 =225 mL.

The standard solution of [tex]HNO_3[/tex] is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles [tex]n=molarity * volume[/tex]

[tex]M_1.V_1=M_2.V_2[/tex]

[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

Explanation:

P1V1/T1 = P2V2/T2

=((390.89×313.9)/76.6)×((718.3×V2)/154.2)

= (122700.371/76.7) × ((718.3×V2)/154.2)

make V2 the subject of the formula...

V2 =(122700.371×154.2)/(76.6×718.3)

V2 =18920397.21/55021.78

V2 = 343.87

Answer:

23592U+10n→14454Xe+9038Sr+210n

Explanation:

a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.

Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

[tex]M=58g[/tex]

Explanation:

From the question we are told that:

Heat Capacity [tex]H=0.897[/tex]

Mass of water [tex]M=200g[/tex]

Initial Temperature of Aluminium [tex]T_a=85.6[/tex]

Initial Temperature of Water [tex]T_{w1}=16.0[/tex]

Final Temperature of Water  [tex]T_{w2}=16.0[/tex]

Generally

Heat loss=Heat Gain

Therefore

[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]

[tex]M=58g[/tex]

In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.

The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:

The addition of barium hydroxide will raise the pH slightly because the buffer still working.

The initial moles of those species are:

Hypochlorous acid:

[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]

Sodium hypochlorite:

[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]

Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:

Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O

For a complete reaction of 0.092 moles of barium hydroxide are required:

[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]

As there are 0.370 moles, the moles of HClO after the reaction are:

0.370 moles - 0.184 moles = 0.186 moles of HClO will remain

As you still have hypochlorite and hypochlorous acid you still have a buffer.

Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.

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Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

D. beta particle

Explanation:

Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.

Hello There!

Hope that helps you!

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[tex]SilentNature[/tex]

Answer:

This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.

Finding Nemo??? sorry I really need these points

Answer:

2-nonyne

explanation:

First consider the type of bonds, there are tripple bonds of carbon to carbon, in position 2 from the right.

hence it us alkyn.

There are 9 carbons.

Answer:

Fe(NO3)3, Cr(NO3)3, Co(NO3)3

Explanation:

According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.

Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.

The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.

The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.

We measured the Fe(II) reduction of one of the Co(III) complexes by water at a rate of about 0.545 kJ/mol (to three significant figures).

Calculating a Reaction's Activation Energy A reaction's rate is influenced by the temperature at which it is carried out. The molecules travel more quickly and clash more frequently as the temperature rises. Moreover, the molecules contain greater kinetic energy.

We can use the Arrhenius equation to calculate the reaction's activation energy:

k = A × exp(-Ea/RT)

When the activation energy Ea, the rate constant k, the gas constant R, and the temperature T in Kelvin are all present.

Finding the natural logarithm of the equation's two sides results in:

ln(k) = ln(A) - (Ea/RT)

This equation can be rearranged to take a linear form:

ln(k) = (-Ea/R) × (1/T) + ln(A)

y = mx + b, where (1/T) is x, (-Ea/R) is the slope, and ln(A) is the y-intercept, has the form of a linear equation.

We can get the slope of the line using the given data:

slope = (-Ea/R) = (ln(k2/k1)) / (1/T2 - 1/T1)

where the rate constants for temperatures T1 and T2, respectively, are k1 and k2.

substituting the specified values:

k1 = 0.054s⁻¹ at 293 K

k2 = 0.100s⁻¹ at 298 K

T1 = 293 K

T2 = 298 K

slope = (-Ea/R)

= (ln(0.100/0.054)) / (1/298 - 1/293)

= 65.5 kJ/mol

Therefore, the activation energy of the reaction is:

Ea = slope * R = 65.5 kJ/mol × 8.314 J/mol-K = 545 J/mol

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Answer:

An atom is the smallest particle of an element that can take part in chemical reaction.

Explanation:

hope it will help u Amri

After the double replacement reaction from mixing 50.0g of sulfuric acid and 40.0 grams of barium chloride is complete, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain.

The equation of the reaction between sulfuric acid and barium chloride is

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

From this equation of reaction, it means 1 mole of barium chloride will completely react with 1 mole of sulfuric acid.

From the question, we have 50.0g of sulfuric acid and 40.0 grams of barium chloride.

First, we will determine the number of moles of the sulfuric acid and barium chloride present.

Number of moles is given by the formula

Number of moles = Mass / Molar mass

For sulfuric acid

Mass = 50.0 g

Molar mass = 98.079 g/mol

∴ Number of moles = 50.0 / 98.079

Numbers of moles of sulfuric = 0.509793 mol

For barium chloride

Mass = 40.0 grams

Molar mass of barium chloride = 208.23 g/mol

∴ Number of moles = 40.0 / 208.23

Number of moles of barium chloride = 0.192095 mol

Since the number of moles of sulfuric acid is more than that of barium chloride, then the limiting reagent is barium chloride and the excess reagent is sulfuric acid

NOTE: A limiting reagent is the reactant that is completely used up in a reaction, and it determines when the reaction stops.

Hence, barium chloride will be used up during the reaction (that is, 0 grams will remain after the reaction is complete).

For the mass of sulfuric acid that will remain,

First, we will determine the number of mole that will remain.

Since 1 mole of barium chloride completely reacts with 1 mole of sulfuric acid, then 0.192095 mol of barium chloride will react with 0.192095 mol of sulfuric acid.

∴ The remaining number moles of sulfuric acid = 0.509793 mol - 0.192095 mol

The remaining number moles of sulfuric acid = 0.317698 mol

Then, from

Mass = Number of moles × Molar mass

Mass = 0.317698 mol ×98.079 g/mol

Mass = 31. 1595 gram

Mass ≅ 31.16 grams

∴ 31.16 grams remains after the reaction is complete.

Hence, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain after the double replacement reaction is complete.

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Answer:

3.310308*10^26

Explanation:

nO=9nFe2(SO3)3=9*60.1=540.9 moles

number of atoms: 540.9*6.02*10^23

Answer:

% NiCl2 = 24.13%

% water = 78.57%

Explanation:

Mass percentage = mass of solute/mass of solution × 100

According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:

Mass of solution = 159g + 500g

Mass of solution = 659g

Therefore;

A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100

% Mass of NiCl2 in solution = 159/659 × 100

% Mass of NiCl2 in solution = 0.2413 × 100

= 24.13%

B) % Mass of water in solution = mass of water/mass of solution × 100

% Mass of water in solution = 500/659 × 100

% Mass of water in solution = 0.7587 × 100

% Mass of water in solution = 75.87%

Answer:

it is 11.55 and ik because I just had that question

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:

[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]

The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:

[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

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Water, mercury chloride and nitrogen oxide.

Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.

https://brainly.com/question/24261598.

Answer:

a) E2

b) SN2

c) SN2

Explanation:

A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.

We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.

Answer:

[tex]K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]

Explanation:

Hello there!

In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:

[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]

Now, when we combine them to get the overall expression, we infer these two are multiplied to get:

[tex]K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]

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Answer:

a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.

Explanation:

Identify the correct descriptions of beta particles.

a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)

b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.

c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.

d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.

Answer:

0

Explanation:

answer from gauth math