HELPPP PLZ !
Which one of the following is an example of a difficulty caused by thermal
expansion?
A. butter melting too fast
B. falling through the ice while skating
C. sagging power lines on a cold day
D. cracking in the walls of a house

Answers

Answer 1
The answer is D. Because thermal expansion makes the wall expand and crack

Related Questions

How many grams are in 5.87 x 10^21 molecules of sulfur?

Answers

Answer:

0.312g

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules

1mole of sulphur = 32g

If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules

Then, Xg of sulphur will contain 5.87x10^21 molecules i.e

Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g

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How would you monitor the progress of a neutralization reaction? Question 2 options: We will use a funnel to separate the solid as it forms We will use a balance to see the changes in mass We will use a thermometer to check the changes in temperature We will use an acid-base indicator to see changes in color depending on the pH

Answers

Answer:

We will use an acid-base indicator to see changes in colour depending on the pH  

Explanation:

The pH changes during a titration, so you could use an acid-base indicator to follow the changes in pH.

A is wrong. An acid-base titration does not usually form a solid, and it would be impractical to isolate a solid with a funnel.

B is wrong. There are no changes in mass.

C is wrong. Any changes in temperature would be too small to measure precisely with an ordinary thermometer.

The best way to monitor the progress of a neutralization reaction such as acid-base titration: D. Use an acid-base indicator to observe the changes in color depending on the pH.

The chemical reaction that occurs when you mix an acid and a base together is referred to as neutralization reaction.

In a neutralization reaction, what is formed is salt and water.

Acid-base titration is a neutralization method.

During acid-base titration, the neutralization reaction that occurs is usually monitored by observing the pH changes that occurs.

Change in pH is an indicator that there is progress in the neutralization reaction.

An acid-base indicator, can be used to detect the changes that occur via the pH changes in relation to the color change.

Therefore, the best way to monitor the progress of a neutralization reaction such as acid-base titration: D. Use an acid-base indicator to observe the changes in color depending on the pH.

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If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?

Answers

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.

What is volume?

Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.

Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.

This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.

Here, it is given that mass of olive oil is 1.2kg.

We know that,

Density of olive oil = 0.917kg/l.

Volume = mass/density

Volume = 1.2/0.917.

Volume = 1.30 lit.

Volume = 1300mL.

Thus, the volume of olive oil will be 1300 mL.

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5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g) 3H 2(g) + N 2(g) At equilibrium and a particular temperature, 1.00 mole of ammonia remains. Calculate K c for the reaction.

Answers

Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)

N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..

~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

J. [,v 1+3] ~

I

4,:i.~ = 0,05

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:

[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]

At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:

[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]

We can calculate the concentrations of all the species at equilibrium using an ICE chart.

       2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)

I          1.00              0          0

C         -2x              +3x        +x

E       1.00-2x          3x          x

Since the concentration of ammonia at equilibrium is 0.200 M,

[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]

The concentrations of all the species at equilibrium are:

[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]

The concentration equilibrium constant (Kc) is:

[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

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Choose the most appropriate indicator for the titration of a weak acid with NaOH, where the expected equivalence point of the titration is at pH 8.8.

a. methyl orange, pH range 3.2-4.4
b. methyl red, pH range 4.8 6.0
c. bromothymol blue, pH range 6.0-7.6
d. phenolphthalein, pH range 8.2-10.0
e. alizarin yellow R. pH range 10.1-12.0

Answers

Answer:

D phenolphthalein,pH range 8.2-10.0

Write the electron configuration when Sulfur gains two electrons

Answers

Answer:

Explanation:

If sulfur gains 2 electrons then two electrons should be added to it electronic configuration.

Arrange the following elements in order of decreasing first ionization energy: S, Ca, F, Rb, and Si.
Rank from largest to smallest. To rank items as equivalent, overlap them.

Answers

Answer:

The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.  

Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.  

Considering the definition of ionization energy,

Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.

You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.

In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.

In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.

Taking into account the above, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.  

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A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?

Answers

Answer:

Molarity of the sulfuric acid solution is 4.61M

Explanation:

The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:

2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻

That means, 2 moles of base react with 1 mole of sulfuric acid.

If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:

0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄

As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:

0.200 moles H₂SO₄ / 0.0434L = 4.61M

Molarity of the sulfuric acid solution is 4.61M

29. Which alcohol combines with carboxylic acid to produce the ester called ethyl butanoate?

A) butan-2-ol

B) propan-1-ol

C) butan-1-ol

D) ethanol

Answers

Answer:

The answer is option D.

ethanol

Hope this helps you

What element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)​

Answers

Answer:

Option A

Explanation:

Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.

Answer:

A. Silicon (Si)

Explanation:

Silicon (Si) is primarily used as a semiconductor material to make electronic chips.

Suppose there is 1.00 L of an aqueous buffer containing 60.0 mmol of formic acid (pKa=3.74) and 40.0 mmol of formate. Calculate the pH of this buffer.

Answers

Answer:

pH = 3.56

Explanation:

The pH of the buffer producing from the mixture of formic acid and formate ion can be found using H-H equation:

pH = pka + log [A⁻] / [HA]

pH = 3.74 + log [Formate] / [formic acid]

Where [] represents molar concentrations -or moles- of formate and formic acid in the solution.

Replacing knowing moles of formic acid = 0.0600 and moles formate = 0.0400:

pH = 3.74 + log [Formate] / [formic acid]

pH = 3.74 + log [0.0400] / [0.0600]

pH = 3.56

1. Define the Law of Conservation of Mass (via text). Now that you’ve defined this law, explain what it means in your own words using an example.

Answers

Explanation:

The law of conservation of mass states that mass can neither be created nor be destroyed.

Explanation in own words = this means that in this universe no one can create or destroy mass.

No physical or chemical force.

How many atoms of hydrogens are found in 3.21 mol of
C3H8?​

Answers

Answer:

1.55 × 10²⁵ atoms of H  

Explanation:

3.21mol C₃H₈ × 8mol H × (6.022×10²³)

Copper was one of the earliest metals used by humans, because it can be prepared from a wide variety of copper minerals, such as cuprite (Cu2O), chalcocite (Cu2S), and malachite [Cu2CO3(OH)2]. Balance the following reactions for converting these minerals into copper metal. Place a coefficient in each gray box.
(a) Cu2O(s) + C(s) rightarrow Cu(s) + CO2(g)
(b) Cu2O(s) + Cu2 S(s) rightarrow Cu(s) + SO2(g)
(c) Cu2 CO3 (OH)2(s) rightarrow CuO(s) + CO2(g) + H2O(g)
Use the left and rightarrow keys to move the cursor out of a superscript or subscript in the module.

Answers

Answer:

a. 2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)

b. 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)

c. Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)

Explanation:

A reaction is balanced when you have the same amount of atoms in reactants and products.

In the reactions:

(a) Cu₂O(s) + C(s) → Cu(s) + CO₂(g)

As a general rule, you first balance oxygen and hydrogen. In products you have 2 oxygens, then:

2 Cu₂O(s) + C(s) → Cu(s) + CO₂(g)

Carbon is balanced yet. Thus, you need just to balance Cu:

2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)

(b) Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)

Balancing oxygen:

2Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)

Sulfur is balanced yet. Now you just need to balance Cu:

2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)

(c) Cu₂CO₃(OH)₂(s) → CuO(s) + CO₂(g) + H₂O(g)

This reaction is different because the reactant is a chemical with a lot of atoms. we will first balance Cu:

Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)

Balancing copper, oxygen, hydrogen and carbon are balanced:

Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)

2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations

Answers

Answer:

a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)

workdone (w) = -8442.6 J  ≈ -8.443 KJ

heat transferred (q) of the ideal gas = - w

q = 8.443 KJ

b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0

the workdone(w) in the ideal gas= - 4567.5 J  ≈ - 4.57 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Explanation:

given

mole of an ideal gas(n) = 2.5 mol

Temperature (T) = 20°C

= (20°C + 273) K  = 293 K

Initial pressure of the ideal gas(P₁) = 20 atm

Final pressure of the ideal gas(P₂) = 5 atm.

2) (a)for adiabatic reversible process,

note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.

Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]

= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]

= 6090.01 J × [-1.3863]

= -8442.6 J  ≈ -8.443 KJ

So, the work done (w) of ideal gas = -8.443 KJ

For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-8.443 KJ)

q = 8.443 KJ

heat transfer (q) of the ideal gas = 8.443 KJ

(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.

Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )

= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]

= - 6090.01 J × 0.75

= - 4567.5 J  ≈ - 4.57 KJ

∴work done(w) of an ideal gas = - 4.57 KJ

For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-4.5675 KJ)

q = 4.5675 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen

Answers

Answer:

15.3 %

Explanation:

Step 1: Given data

Mass of the sample (ms): 230 gMass of carbon (mC); 136.6 gMass of hydrogen (mH): 26.4 gMass of nitrogen (mN): 31.8 g

Step 2: Calculate the mass of oxygen (mO)

The mass of the sample is equal to the sum of the masses of all the elements.

ms = mC + mH + mN + mO

mO = ms - mC - mH - mN

mO = 230 g - 136.6 g - 26.4 g - 31.8 g

mO = 35.2 g

Step 3: Calculate the mass percent of oxygen

%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %

Compare strontium with rubidium in terms of the following properties:
a. Atomic radius, number of valence electrons, ionization energy.
b. Strontium is smaller than rubidium.
c. Rubidium is smaller than strontium.
d. Strontium has more valence electrons.
e. Rubidium has more valence electrons.
f. Strontium has a larger ionization energy.
g. Rubidium has a larger ionization energy.

Answers

Answer:

Strontium is smaller

Strontium has the higher ionization energy

Strontium has more valence electrons

Explanation:

It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table

While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)

Since they are in the same period, periodic trends would be useful in evaluating their properties

In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size

Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius

Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has

In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case

4.2 mol of oxygen and 4.0 mol of NO are introduced to an evacuated 0.50 L reaction vessel. At a specific temperature, the equilibrium 2NO(g) + O 2(g) 2NO 2(g) is reached when [NO] = 1.6 M. Calculate K c for the reaction at this temperature.

Answers

Explanation:

At 593K a particular decomposition’s rate constant had a value of 5.21×10−4 and at 673K the same reaction’s rate constant was 7.42×10−3. It was noticed that when the reactant’s initial concentration was 0.2264 M (with a 593K reaction temperature), the initial reaction rate was identical to the initial rate when the decomposition was run at 673K with an initial reactant concentration of 0.05999 M. Recall that rate laws have the form rate = k [A]x and, showing work, determine the order of the decomposition reaction.

Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. Use the theoretical yield of liquid iron and the mass or iron ingots to calculate the percent yield of the reaction.

Answers

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles[/tex]

[tex]\text{Moles of} CO=\frac{260}{28}=9.3moles[/tex]

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]

According to stoichiometry :

1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex]  of [tex]CO[/tex]

Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.

As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]

Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

[tex]\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%[/tex]

Therefore, the percent yield is, 91.8%

A 45.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH.? Determine the pH of the solution after adding 35.0 mL of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + NaOH (aq) D NaC2H3O2(aq) + H2O (l) (Hint: Calculate new concentration and ICE table)

Answers

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

.02M

CH₃COOH  = CH₃COO⁻ + H⁺

C                       xC             xC

Ka = xC . xC / C = x² C

1.8 x 10⁻⁵ = x² . .02

x² = 9 x 10⁻⁴

x = 3 x 10⁻²

= .03

concentration of H⁺ = xC = .03 . .02

= 6 x 10⁻⁴ M , volume =  45 x 10⁻³ L

moles of H⁺  = 6 X 10⁻⁴  x 45 x 10⁻³

= 270 x 10⁻⁷ moles

= 2.7 x 10⁻⁵ moles

concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L

moles of Na OH = 2 X 10⁻²  x 35 x 10⁻³

= 70 x 10⁻⁵ moles

=  

NaOH is a strong base so it will dissociate fully .

there will be neutralisation reaction between the two .

Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles

= 67.3 x 10⁻⁵ moles of NaOH

Total volume = 45 + 35 = 80 x 10⁻³

concentration of NaOH after neutralisation.= 67.3  x 10⁻⁵ / 80 x 10⁻³ moles / L

= 8.4125  x 10⁻³ moles / L

OH⁻ = 8.4125  x 10⁻³

H⁺ = 10⁻¹⁴ / 8.4125  x 10⁻³

= 1.1887 x 10⁻¹²

pH = - log (  1.1887 x 10⁻¹² )

= 12 - log 1.1887

= 12 - .075

= 11.925 .

How many Liters of 0.968M solution can be made if 0.581 moles of solute are added? Group of answer choices 0.600 L 60 mL 0.562 L 1.00 L

Answers

Answer:

0.6L

Explanation:

The formula of molarity is molSolute/litreSolution

[tex]0.968M=\frac{0.581}{LitreSolution} \\\\LitreSolution=\frac{0.581}{0.968} \\LitreSolution=0.6L[/tex].

For the imine synthesis reaction, the two reactants react in melting state. How is that possible since the melting points for both ortho-vanilin and para-toluidine are above room temperature

Answers

Answer:

I have no clue at all im in 11 and dont know anything lol byeeee

Explanation:um um i am lost

Use 1-Butanol as the only organic compound, design a method to synthesize 5-Nonanone. You may use any other inorganic reagents. Any organic reagents have to be made from 1-butanol.

Answers

Answer:

See attached picture.

Explanation:

Hello,

In this case, starting by 1-butanol, we can make it react with hydrogen bromide (1st step) in order to yield 1-bromobutane. Next, the formed alkyl halide is treated magnesium in the presence of an ether in order to yield butyl magnesium bromide which is a Grignard reagent (2nd step). Finally, by adding carbon dioxide, water and extra hydrogen chloride, a carbonyl group can be formed between two butyl radicals in order to form the 5-nonanone (3rd step) as shown on the attached picture.

Best regards.

In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 176 kg/h. Determine the percentage of excess air used during this process.

Answers

Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

Which types of electron orbitals will have higher energy than a 4d orbital?
A. 4p
B. 3s
C. 4f
D. 5s​

Answers

5s orbitals will have higher energy

Answer:

D: 5s

Explanation:

hope this helps :)

What does the period number tell about the energy levels occupied by
electrons in an atom?
A. The period number tells how many electrons are in the highest
energy level of the atom.
B. The period number tells which is the highest energy level occupied
by the electrons.
C. The period number tells how many electrons are in each sublevel
of the atom.
D. The period number tells how many energy sublevels are occupied
in the atom.

Answers

Answer: B. The period number tells which is the highest energy level occupied by the electrons

Explanation:

The period number ( denoted by 'n' ) is the outer energy level that is occupied by electrons in an atom. The period number that an element is in, is the number of energy levels that the element has.When we move across a period from left to right in a periodic table the number of electrons in atoms increases within the same orbit.

Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.

hence, the correct option is B. The period number tells which is the highest energy level occupied  by the electrons.

The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.

What are energy levels ?

The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.

When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.

Therefore, option B , second option is correct.

Learn more about   energy levels at;

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Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is

Answers

Answer:

The mass of NaHCO3 required is 235.22 g

Explanation:

*******

Continuation of Question:

2NaHCO3(s) + H2SO4(aq)  →  Na2SO4(aq) + 2CO2(g) + 2H2O(l)

You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?

********\

The question requires us to calculate the mass of NaHCO3  to neutralize the acid.

From the balanced chemical equation;

1 mol of H2SO4 requires 2 mol of NaHCO3

1.4 would require x?

Upon solving for x we have;

x = 1.4 * 2 = 2.8 mol of NaHCO3

The relationship between mass and number of moles is given as;

Mass = Number of moles * Molar mass

Mass = 2.8 mol * 84.007 g/mol

Mass =  235.22 g

Why is it important that the primary standard chemical be non-hygroscopic and pure? Why is it important to dry the primary standard to a constant weight?

Answers

Answer:

It is extremely important for the primary standard chemical to be non – hygroscopic and pure and to also have a constant weight because you don't want any moisture or any impurities to alter the stoichiometric point in the reaction

It is important that the primary standard chemical be non-hygroscopic and pure to calculate the exact calculation of the reaction.

What is non hygroscopic chemicals?

Non hygroscopic chemicals are those compounds which will not absorb water or mositure from the outside.

If we take any substance which are hygroscopic in nature and during the chemical reaction if they absorb water content or moisture then the mass of that substance will alter and changes all the calculation of the reaction.

So, to maintain the stability of calculation we use non hygroscopic materials.

To know more about non hygroscopic materials, visit the below link:

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A________ chemical bond is a strong attraction between two or more atoms.

Answers

Answer:

Covalent bond is a chemical bond is a strong attraction between two or more atoms.

Answer:

Ionic

Explanation:

An ionic chemical bond is a strong attraction between two or more atoms. Two atoms are more strongly attracted by ionic bonds.

Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water and carbon dioxide chemically react to form the simple sugar glucose and oxygen gas . What mass of simple sugar glucose is produced by the reaction of 4.9 of carbon dioxide?

Answers

Answer:

3.3 g of glucose, C6H12O6.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

6CO2 + 6H2O —> C6H12O6 + 6O2

Next, we shall determine the mass of CO2 that reacted and the mass of C6H12O6 produced from the balanced equation.

This is illustrated below:

Molar mass of CO2 = 12 + (2x16) = 44 g/mol

Mass of CO2 from the balanced equation = 6 x 44 = 264 g

Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180 g/mol

Mass of C6H12O6 from the balanced equation = 1 x 180 = 180 g

From the balanced equation above,

264 g of CO2 reacted to produce 180 g of C6H12O6.

Finally, we shall determine the mass of C6H12O6 produced by reacting 4.9 g of CO2 as follow:

From the balanced equation above,

264 g of CO2 reacted to produce 180 g of C6H12O6.

Therefore, 4.9 g of CO2 will react to produce = (4.9 x 180)/264 = 3.3 g of C6H12O6.

Therefore, 3.3 g of glucose, C6H12O6 were obtained from the reaction.

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