A student found the mass of an object to be 26.5 g. To find the volume, the student submerged the object in a graduated cylinder of water. Submerging the object in the water in the graduated cylinder increased the water level by 24.1 mL. The density of the object is
Question 17 options:

A) 0.909 g/mL.

B) 1.1 g/mL.

C) 1.10 g/mL.

D) 1.0906 g/mL.

Answer:

A. 10.29 g of C2H6.

B. C2H4.

C. 3.14 g of H2.

Explanation:

Step 1:

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

C2H4 + H2 —> C2H6

Step 2:

Determination of the masses of C2H4 and H2 that reacted and the mass of C2H6 produced from the balanced equation.

This is illustrated below:

Molar mass of C2H4 = (12x2) + (4x1) =28 g/mol

Mass of C2H4 from the balanced equation = 1 x 28 = 28 g.

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 1 x 2 = 2 g.

Molar mass of C2H6 = (12x2) + (6x1) = 30 g/mol.

Mass of C2H6 from the balanced equation = 1 x 30 = 30 g

From the balanced equation above,

28 g of C2H4 reacted with 2 g of H2 to produce 30 g of C2H6.

Step 3:

Determination of the limiting reactant. This can be obtained as follow:

From the balanced equation above,

28 g of C2H4 reacted with 2 g of H2.

Therefore, 9.6 g of C2H4 will react with = (9.6 x 2)/28 = 0.69 g of H2.

From the calculations made above, we can see that only 0.69 g out of 3.83 g of H2 given is needed to react completely with 9.6 g of C2H4.

Therefore, C2H4 is the limiting reactant and H2 is the excess reactant.

A. Determination of the maximum mass of ethane, C2H6 produced from the react.

In this case, the limiting reactant will be used because it will produce the maximum yield of the reaction since all of it is consumed in the reaction.

The limiting reactant is C2H4 and the maximum mass of C2H6 can be obtained as follow:

From the balanced equation above,

28 g of C2H4 reacted to produce 30 g of C2H6.

Therefore, 9.6 gof C2H4 will react to produce = (9.6 x 30)/28 = 10.29 g of C2H6.

Therefore, 10.29 g of ethane, C2H6 were produced from the reaction.

B. The limiting reactant is ethylene with formula C2H4. Please refer to step 3 above for details.

C . Determination of the mass of the excess reactant that remained after the reaction.

The excess reactant is H2, please refer to step 3 above for details and the mass that remained after the reaction can be obtained as follow:

Mass of H2 given = 3.83 g

Mass of H2 that reacted = 0.69 g

Mass of H2 remaining =.?

Mass of H2 remaining = mass of H2 given – mass of H2 that reacted.

Mass of H2 remaining = 3.83 – 0.69

Mass of H2 remaining = 3.14 g

Therefore, 3.14 g of H2 remained after the reaction.

Answer:

15.0L

Explanation:

p/v = constan

(9*21)/253 =(15v)/ 302

v = (9*21*302)/(15*253)

v=15.0

Explanation:

I hope it helps you

good luck

Answer for the question

Answer:

0,

Explanation:

if n was 2, then 1,0,-1

Answer:

Helium

Explanation:

Though all noble gases are stable, helium only has two electrons and one shell so it basically doesn't have room for others.

Answer:

[C₆H₅NH₃⁺] = 0.0399 M

Explanation:

This excersise can be easily solved by the Henderson Hasselbach equation

C₆H₅NH₃Cl → C₆H₅NH₃⁺  + Cl⁻

pOH = pKb + log (salt/base)

As we have value of pH, we need to determine the pOH

14 - pH = pOH

pOH = 8.43  (14 - 5.57)

Now we replace data:

pOH = pKb + log ( C₆H₅NH₃⁺/  C₆H₅NH₂ )

8.43 = 9.13 + log (  C₆H₅NH₃⁺ / 0.2 )

-0.7 = log (  C₆H₅NH₃⁺ / 0.2 )

10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2

0.19952 = C₆H₅NH₃⁺ / 0.2

C₆H₅NH₃⁺ = 0.19952 . 0.2  = 0.0399 M

Answer: E = - 19.611×[tex]10^{-18}[/tex] J

Explanation: The lowest possible energy can be calculated using the formula:

[tex]E_{n} = - Z^{2}.\frac{k}{n^{2}}[/tex]

where:

Z is atomic number of the atom;

k is a constant which contains other constants and is 2.179×[tex]10^{-18}[/tex] J

n is a layer;

For the lowest possible, n=1.

Atom of Lithium has atomic number of Z=3

Substituing:

[tex]E_{1} = - 3^{2}.\frac{2.179.10^{-18}}{1}[/tex]

[tex]E_{1} =[/tex] [tex]-19.611.10^{-18}[/tex] J

The energy for the electron in the [tex]Li^{+2}[/tex] ion is - 19.611 joules

The lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion is equal to [tex]1.96\times 10^{-17}\; Joules[/tex]

To determine the lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion, we would use the Bohr model:

Mathematically, Bohr's model is given by the equation:

[tex]Energy = -Z^2 \frac{k}{n^2}[/tex]

Where:

We know that the atomic number of lithium (Li) is equal to 3.

Also, at the lowest possible energy, n = 1.

Rydberg constant = [tex]2.179 \times 10^{-18}[/tex]

Substituting the parameters into the equation, we have;

[tex]E_1 = -3^2 \times \frac{2.179 \times 10^{-18}}{1^2} \\\\E_1 =9 \times 2.179 \times 10^{-18}\\\\E_1 =1.96\times 10^{-17}\; Joules[/tex]

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Answer:

Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Look up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].

The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.

How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?

[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].

There are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

The molarity of a solution measures its molar concentration. For this solution:

[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

(Rounded to four significant figures.)

Answer:

-285.4 J/K

Explanation:

Let's consider the following balanced equation.

HCl(g) + NH₃(g) ⇒ NH₄Cl(s)

We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.

ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))

ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol

ΔS°r = -285.4 J/K

Answer:

-198.3 J/K mol

Explanation:

I got it correct on founders edtell

Answer:

Total ATP molecules produced = 66 molecules of ATP

Explanation:

A 10-carbon fatty acid when it has undergone complete oxidation will yield 5 acetyl-CoA molecules and 4 FADH₂ and 4 NADH molecules each. Each of the 5 acetyl-CoA molecules enters into the citric acid cycle and is completely oxidized to yield further ATP and  FADH₂ and NADH molecules.

The total yield of ATP in the various enzymatic step is calculated below:

Acyl-CoA dehydrodenase = 4 FADH₂

β-Hydroxyacyl-CoA dehydrogenase = 4 NADH

Isocitrate dehydrogenase = 5 NADH

α-Ketoglutarate dehydrogenase = 5 NADH

Succinyl-CoA synthase = 5 ATP (from substrate-level phosphorylation of GDP)

Succinate dehydrogenase = 5 FADH₂

Malate dehydrogenase = 5 NADH

Total ATP  from FADH₂ molecoles = 9 * 1.5 = 13.5

Total NADH molecules = 19 * 2.5 = 47.5

Total ATP molecules produced = 13.5 + 47.5 + 5

Total ATP molecules produced = 66 molecules of ATP

Answer:

Explanation:

First, calculate the number of acetyl-CoA molecules formed:

Number of acetyl-CoA molecules = [tex]\frac{number of carbons in fatty acid}{2}[/tex]

[tex]= \frac{10}{2}\\\\ = 5 acetyl-CoA molecules[/tex]

Next, calculate the number of rounds of beta-oxidation:

Number of rounds = number of acetyl-CoA molecules - 1

[tex]= 5 - 1\\\\ = 4 rounds[/tex]

Calculate the number of ATP from NADH and FADH2:

If each NADH yields 2.5 ATPs and each FADH2 yields 1.5 ATPs, then multiply the number of rounds by 4 and multiply the number of acetyl-CoA molecules by 10.

[tex](4 * 4) + (5 * 10) = 66 ATP[/tex]

Subract two ATP molecules for activation of the fatty acid.

[tex]Total ATP = 66 - 2\\\\ = 64 ATPs[/tex]

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Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

Answer:

6.022 × 10^23

hope this helps :)

Answer:

Replacing 3 of the H of LiALH4 with OR groups make it a more reactive reducing agents than NaBH4

Explanation:

LiAlH4 and NaBH4 are two well known reducing agents in organic chemistry. These two reducing agents function by transfer of hydrogen to the substrate.

If the hydrogen atoms in LiAlH4 are replaced by the -OR moiety, the new compound will be far less reducing than NaBH4 because the hydrogen atoms necessary to effect the reduction has been removed. Thus, the new compound containing -OR moiety can never be more reducing than NaBH4. This implies that the statement written in the answer is false as written.

Answer:

Graphite> sodium chloride> solid ammonia

Explanation:

Melting points of solids has a lot to do with the nature of intermolecular forces in the solid. A substance melts when the intermolecular forces holding the crystal lattice has been overcome such that that the crystal structure of the solid just collapses.

Graphite consists of covalently bonded layers of carbon atom which form a giant lattice. The melting point of graphite is very high because of the fact that the strong covalent bonds that hold the carbon atoms together in the layers require a lot of heat energy to break. Grapoghite melts at about 3600°C

Sodium chloride is an ionic compound that melts at about 801°C. The lattice is composed of alternate sodium and chloride ions.

Solid ammonia is held together by much weaker intermolecular interaction hence it has a melting point of about −77.73 °C.

Answer:

- Pressure in St. Paul, Minnesota

- Temperature in St. Paul, Minnesota

Explanation:

22.4 L or dm³ is the volume for a gas under Standard pressure and temperature conditions.

It is logically to say, that tempereature value at the day of the experiment was not 273.15 K, which is 32°F

We can say, that the pressure was not 1 atm. St Paul Minnesota has  a minimum, but a little height, so the pressure differs by few figures from the standard pressure values.

We also have to mention, that 22.4 L is the value for the Ideal gases at standards conditions. Ideal gases does not exisist on practice, we always talk about real gases. Don't forget the Ideal Gases Law equation:

P . V = n . R . T

Pressure . Volume = number of moles . 0.082 L.atm /mol. K  . 273.15K

Number of moles must be 1 at STP, to determine a volume of 22.4L

The name of the molecule which is given below is 2-pentene.

Alkenes are the organic compounds which are composed of carbon and hydrogen atoms, in which double bond is present.

In the given diagram:

So the compound is 2 pentene.

Hence, 2 pentene is the name of the compound.

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Answer:

O, S, Te

Cl, Br, Se

Explanation:

Main group elements have an electronegativity that increases across a period (from left to right) and decreases down a group.

Each atom has its own value which you can find on the electronegativity chart.

Metals have low values

Nonmetals have high values

I'm your case:

O = 3.5

S = 2.5

Te = 2.1

Cl = 3.0

Br = 2.8

Se = 2.4

O, S, Te, and Cl, Br, Se are the correct order for the elements listed in order of decreasing electronegativity.

Electronegativity. is the tendency or the efficiency of an atom to attract the lone pair of electrons towards itself to become stable and complete their octave is known as electronegativity.

In the periodic table oxygen, fluorine and nitrogen are the three top, most electronegative elements, and from moving up to down in periodic table electronegativity. decreases and left to right increases.

Therefore,  the correct order for the elements listed in order of decreasing electronegativity is O, S, Te, and Cl, Br, Se.

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Answer:

b) 1.40 V

Explanation:

Oxidation half equation;

2Fe(s) → 2Fe3+(aq) + 6 e-

Reduction half equation;

3Cl2(g) + 6 e- → 6 Cl-(aq)

E°cathode= 1.36 V

E°anode= -0.036 V

E°cell= E°cathode - E°anode

E°cell= 1.36 -(-0.036)

E°cell= 1.36 + 0.036

E°cell= 1.396 V

E°cell= 1.40 V

Answer:

The equilbrium concentration of NH₃(g) is 2.4 M

Explanation:

The balanced reaction is:

4 NH₃(g) + 7 O₂(g) ⇔ 2 N₂O₄(g) + 6 H₂O(g)

By stoichiometry of the reaction,  2 moles of N₂O₄ are formed from 4 moles of NH₃.

Considering that the concentration is [tex]concentration=\frac{number of moles}{volume}[/tex] and with a volume of 1 liter, it is possible to apply the following rule of three: if 2 M of N₂O₄ are formed from 4 M of NH₃, 0.6 M of N₂O₄ from what concentration  of NH₃ are formed?

[tex]concentration of NH_{3}=\frac{0.6 M of N_{2}O_{3} *4MofNH_{3} }{2 M of N_{2}O_{3} }[/tex]

concentration of NH₃= 1.2 M

By subtracting the moles of NH3 in equilibrium from the moles of NH₃ initially, you will see how many moles of NH₃ were converted and remain in equilibrium: 3.6 M - 1.2 M= 2.4 M

The equilbrium concentration of NH₃(g) is 2.4 M

simple distillation can be used when the temperature difference between the boiling points of two miscible liquid is at least 25°c. the temperature difference between the boiling points of kerosene and petrol is 25c. hence, this mixture can separated using simple distillation.

simple distillation

Answer:

d, 34 neutrons

Explanation:

Mass number is the sum of the number of protons and the number of neutrons.

Atomic number equals to the number of protons.

Since the element has an atomic number of 31, it has 31 protons. The number of neutrons equal to mass number - no. of protons,

which in this case is 65 - 31

= 34 neutrons.

In a neutral atom, the no. of electrons equal to the no. of protons, because each proton carries a +1 charge and each  electron carries a -1 charge. To cancel out the charge, their numbers must be equal. So, this atom has 31 electrons.

From the options, only d is correct. (34 neutrons)

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

[tex]\frac{3}{1} =\frac{y}{0.02256} \\[/tex]

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

Answer:

Molarity of sodium acetate you will need to add is 0.0324M

Explanation:

Assuming volume of the buffer is 1L.

The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is pKa of the weak acid,  [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid

Replacing for the acetic buffer (pKa = 4.76):

pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]

As you have 0.010 moles of acetic acid in 1L:

[Acetic Acid] = 0.010mol / 1L = 0.010M

And you require a pH of 5.27:

5.27 = 4.76 + log [Sodium Acetate] / [0.010M]

0.51 = log [Sodium Acetate] / [0.010M]

10^0.51 = [Sodium Acetate] / [0.010M]

3.236 =  [Sodium Acetate] / [0.010M]

3.236 [0.010M] = [Sodium Acetate]

0.0324M = [Sodium Acetate]

Answer:

Helium and potassium

Explanation:

The density of helium is 0.18 and potassium 0.86, while lithium is 0.53

Answer:

pH = 3.23

Explanation:

Before the addition of any NaOH, the only you have is a 0.020M acetic acid solution. That is in equilibrium with water as follows:

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

The Ka of this reaction is:

Ka = 1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

Where [] are concentrations in equilibrium of each species

As you have in solution just HC₂H₃O₂, the equilibrium concentrations will be:

[HC₂H₃O₂] = 0.020M - X

[C₂H₃O₂⁻] = X

[H₃O⁺] = X

Where X is reaction coordinate.

Repalcing in Ka expression:

1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]

1.8x10⁻⁵ = [X] [X] / [0.020M - X]

3.6x10⁻⁷ - 1.8x10⁻⁵X = X²

3.6x10⁻⁷ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.0006M → False solution. There is no negative concentrations

X = 0.000591M → Right solution.

As:

[H₃O⁺] = X

[H₃O⁺] = 0.000591M

As pH = -log[H₃O⁺]

Answer:B

Explanation: A P E X

Answer:

C. Rows 6 and 7, separated from the rest of the table

Explanation:

The lanthanides and actinides are groups of elements in the periodic table, that are thirty (30) in number. They are separated from the rest of the periodic table, usually appearing as separate rows at the bottom. They are often called the inner transition metals, because they all fill the f-block.

Therefore, the correct option is C

" They are found in Rows 6 and 7, separated from the rest of the table"

Answer: The average atomic mass of oxygen is 15.999 amu

Explanation:

Mass of isotope O-16 = 15.995 amu

% abundance of isotope O-16= 99.759 % = [tex]\frac{99.759}{100}=0.99759[/tex]

Mass of isotope O-17 = 16.995 amu

% abundance of isotope O-17 = 0.037% = [tex]\frac{0.037}{100}=0.00037[/tex]

Mass of isotope O-18 = 17.999 amu

% abundance of isotope O-18 = 0.204% = [tex]\frac{0.204}{100}=0.00204[/tex]

Formula used for average atomic mass of an element :

[tex]\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})[/tex]

[tex]A=\sum[(15.995\times 0.99759)+(16.995\times 0.00037)+(17.999 \times 0.00204)][/tex]

[tex]A=15.999[/tex]

Thus the average atomic mass of oxygen is 15.999 amu

Answer:

Converting the percent abundance into decimal form, we get:

O-16: 99.759% = 99.759/100 = 0.9975

O-17: 0.037% = 0.037/100 = 0.00037

O-18: 0.204% = 0.204/100 = 0.0020

Average atomic mass of oxygen is:

(15.995) x (0.9975) + (16.995) x (0.00037) + (17.999) x (0.0020)

= 15.955 + 0.0062 + 0.0359

= 15.997 amu

Explanation:

From PLATO

Answer:

26.5 kD  

Explanation:

Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.56 - ( given ). This is only one part of the information we are given / can conclude in this case ....

i = van’t Hoff factor = 1 for a protein molecule,

R = gas constant = 62.36 L torr / K-mol,

T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K

( Known initially ) ∏ = osmotic pressure = 0.56 torr

..... besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.

_____

Substitute derived / known values to solve for M ( moles / liter ) -

∏ = iMRT

⇒ 0.56 = ( 1 )( M )( 62.36 )( 298 )

⇒ 0.56 = M( 18583.28 )

⇒ M = 0.56 / 18583.28 ≈ 0.00003013461 ....

_____

We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -

M = mol / l

⇒ 0.00003013461 = 0.020 / 25 mL ( 0.025 L ),

0.020 / 0.025 = 0.8 g / L

⇒ 0.8 g = 0.00003013461 moles,

molar mass = 0.8 g / 0.00003013461 moles = 26,548 g / mol = 26.5 kD  

Answer:

Solution A is a Weak Alkali, Solution B is a strong Acid.

Explanation:

At pH 10, the colour is blue, therefore it's a weak alkali.

At pH 1, the colour is red, therefore it's a strong Acid.