What is the cost of 4 oranges, if 6 oranges weigh 2.4 pounds and the price of oranges are
$0.50 per pound?

Answer:

59.16 g

Explanation:

Mass = Density × Volume

= 3.4 g/mL × 17.4 mL

= 59.16 g or

59.2 g (rounded to three significant figures) or

59 g (rounded to two significant figures)

Hope that helps.

The mass of a 17.4 mL piece of material, if the density is 3.4 g/mL is 59.16 grams.

Density is defined as the degree to which a material is packed together.

It can also be defined as a material is defined as its total mass (m) divided by its total volume (V).

Density is an important topic because it tells us which compounds will float and which will sink when placed in a liquid. Usually substances float as long as their density is less than that of the liquid in which they are immersed.

Density can be calculated in kilogram per cubic meter or gram per cubic meter.

Density can be expressed as

Density = mass / volume

Given Volume = 17.4 ml

           Density = 3.4 g / ml

Mass = density x volume

Mass = 3.4 x 17.4

Mass = 59.16 grams

Thus, the mass of a 17.4 mL piece of material, if the density is 3.4 g/mL is 59.16 grams.

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Answer:

40

Explanation:

Your trying to find out the meters so your going to divide 3920J by 10 and 9.8

3920/10/9.8

Answer:

Keep a close eye on her in case she starts giving birth to the next pup at the same time. A greenish/brown discharge may suggest a placenta has separated. If you see this, a puppy should be born within the next 2-4 hours.

Answer:

The correct answer is 1.06 M

Step-by-step explanation:

We have to calculate the molarity (M), which is:

M= moles solute/ 1 L solution.

The chemical formulae of manganese (II) nitrate is Mn(NO₃)₂. So, we first calculate its molecular weight (Mw) as follows:

Mw(Mn(NO₃)₂)= molar mass Mn + (2 x molar mass N) + (6 x molar mass N)= 55 g/mol + (2 x 14 g/mol) + (6 x 16 g/mol) = 179 g/mol

Then, with Mw we calculate the number of moles there is in the given mass of Mn(NO₃)₂:

moles Mn(NO₃)₂= mass/Mw= 23.8 g/(179 g/mol)= 0.133 mol

Now, we need the final volume in liters, so we convert the volume from mL to L:

125 mL x 1 L/1000 mL = 0.125 L

Finally, we divide the moles of Mn(NO₃)₂ into the volume in L, to obtain the molarity in mol/L:

M= 0.133 moles/0.125 L = 1.06 mol/L= 1.06 M

Answer:

20 drops

Explanation:

Step 1: Given data

Step 2: Calculate the volume of water added

We calculate the volume of water added by subtracting the initial volume to the final volume.

36.0 mL - 23.0 mL = 13.0 mL

Step 3: Calculate the number of drops in 1 mL

We will divide the number of drops of water added by the volume they represent.

265 drops / 13.0 mL = 20.4 drops/1 mL ≈ 20 drops/1 mL

Answer:

B- three 1H signals and four 13C signals

Explanation:

Protons in similar environment have the same chemical shift; deshielding by electro-negative elements is also a factor that plays in signal emission in spectroscopy.

All C- atoms in methyl-3 group would count as a single peak, as well as the hydrogen atoms present there on the 13C and 1H spectra respectively.

Let me know if you have any further questions.

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

Answer:

Solubility = [SO₄²⁻]

Explanation:

Solubility of Ag₂SO₄ can be understood seen its Ksp equilibrium:

Ag₂SO₄(s) ⇄ SO₄²⁻ + 2Ag⁺

Where 1 mole of silver sulfate dissolves producing 1 mole of sulfate ion and 2 moles of silver ion.

Solubility is defined as: "the amount of solid that can be dissolved per liter of solution".

In the dissolution process, the moles of Ag₂SO₄ dissolved are equal to moles of SO₄²⁻.

That means:

Answer:

4.02 gallons

Explanation:

From the illustration:

$0.750 USD = $1.00 CAD

Therefore,

$20 USD =  20 x 1/0.750 = $26.667 CAD

The cost of gas in the US is 2.21 USD per gallon.

20 USD will be able to get 20/2.21 = 9.05 gallons

The cost of gas across the border is 1.40 CAD per liter.

0.264 gallon = 1 liter

1 gallon = 1/0.264 = 3.788 liters

This means that 1 gallon = 3.788 x 1.40 = 5.303 CAD

Hence, 26.667 CAD will get 26.667/5.303 = 5.03 gallons

If he buys a $20 USD gas before crossing the border, Carl will get 9.05 gallons, but if he buys it after crossing the border, he will get 5.03 gallons. Therefore, if he buys the gas before crossing the border instead of buying it in Canada, he will get 9.05 - 5.03 = 4.02 gallons extra.

Answer:

The volume is 4.13793 L

Explanation:

Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:

[tex]density=\frac{mass}{volume}[/tex]

Density is a characteristic property of every body or substance.

The most commonly used units of density are [tex]\frac{kg}{m^{3} }[/tex] or [tex]\frac{g}{cm^{3} }[/tex] for solids, and [tex]\frac{kg}{L}[/tex] or [tex]\frac{g}{mL}[/tex] for liquids and gases.

In this case, you know:

Replacing:

[tex]0.87\frac{g}{mL} =\frac{3,600 g}{volume}[/tex]

Solving:

[tex]volume =\frac{3,600 g}{0.87\frac{g}{mL}}[/tex]

volume= 4,137.93 mL

Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L

The volume is 4.13793 L

Answer:

Option e.

Acid oxalate / oxalate

Explanation:

To determine the most efecttive buffer, we must look the pKas of the system.

If pKa value is similiar to buffer's pH, the variation of pH by  the addition of a small amount of strong acid or base is minimal. It would be a good buffer.

a. H₃PO₄ / H₂PO₄⁻   Ka 1 = 6.9×10⁻³

pKa = - log Ka → - log 6.9×10⁻³ = 2.16

b. H₂CO₃ / HCO₃⁻    Ka1 = 4.8×10⁻⁷

pKa = - log Ka → - log 4.8×10⁻⁷ = 6.32

c. HCO₃⁻ / CO₃⁻²    Ka2 = 4.8×10⁻¹¹

pKa = - log Ka → - log 4.8×10⁻¹¹ =  10.32

d.  H₂S / HS⁻    Ka1 = 8.9×10⁻⁸

pKa = - log Ka → - log 8.9×10⁻⁸ = 7.05

e. HC₂O₄⁻ / C₂O₄⁻²     Ka2 = 5.1×10⁻⁵

pKa = - log Ka → - log 5.1×10⁻⁵ = 4.29

pKa which is so close to the ph's value is e, acid oxalate / oxalate

Answer:

[tex]M=4.00\frac{mol}{L}=4M[/tex]

Explanation:

Hello.

In this case, we should remember that a molar concentration is defined in terms of the moles and volume in liters as shown below:

[tex]M=\frac{mol}{L}[/tex]

Thus, for the given information we obtain:

[tex]M=\frac{8.000mol}{2.00L}\\ \\M=4.00\frac{mol}{L}=4M[/tex]

Best regards.

The question is missing parts. The complete question is as follows.

Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:

[tex]SO_{2}_{(g)} + \frac{1}{2}O_{2}[/tex] ⇔ [tex]SO_{3}_{(g)}[/tex]; [tex]K_{1}[/tex]

[tex]2SO_{3}_{(g)}[/tex] ⇔ [tex]2SO_{2}_{(g)}+O_{2}_{(g)}; K_{2}[/tex]

The values of the equilibrium constants are related by:

a) [tex]K_{1}[/tex] = [tex]K_{2}[/tex]

b) [tex]K_{2} = K_{1}^{2}[/tex]

c) [tex]K_{2} = \frac{1}{K_{1}^{2}}[/tex]

d) [tex]K_{2}=\frac{1}{K_{1}}[/tex]

Answer: c) [tex]K_{2} = \frac{1}{K_{1}^{2}}[/tex]

Explanation: Equilibrium constant is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:

[tex]K=\frac{[products]^{n}}{[reagents]^{m}}[/tex]

The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.

The equilibrium constants of each reaction are:

[tex]SO_{2}_{(g)} + \frac{1}{2}O_{2}[/tex] ⇔ [tex]SO_{3}_{(g)}[/tex]

[tex]K_{1}=\frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}[/tex]

[tex]2SO_{3}_{(g)}[/tex] ⇔ [tex]2SO_{2}_{(g)}+O_{2}_{(g)}[/tex]

[tex]K_{2}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}[/tex]

Now, analysing each constant, it is easy to see that [tex]K_{1}[/tex] is the inverse of [tex]K_{2}[/tex].

If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:

[tex]K_{2}=\frac{1}{K_{1}^{2}}[/tex]

Answer:

The specific heat capacity is the heat that a body or a system needs to administer so that it can increase its internal temperature.

Explanation:

The calorific capacity is measured in several units, it varies a lot between products, reactants or the same systems since each one is independent in its composition and this conditions it.

As for its mathematical calculation, it is the quotient, that is, the division between the dose of energy transferred to a body and the change in temperature that it experiences.

Answer: 7.71 + 4.31 + 1.7 + 4.00141= 17.72141

7.71 x 4.31 x 1.7 x 4.00141=  226.04433255

Explanation:

(i) 7.71 + 4.31 + 1.7 + 4.00141

= 7.71000+4.31000+1.70000+ 4.00141   [make like decimals]

= 17.72141    [By adding the corresponding places ]

(ii)  7.71 x 4.31 x 1.7 x 4.00141 =( 7.71 x 4.31) x 1.7 x 4.00141

= 33.2301 x 1.7 x 4.00141

= (33.2301 x 1.7) x 4.00141

=  56.49117 x 4.00141

= ( 56.49117 x 4.00141 )

=  226.04433255

Hence,  7.71 + 4.31 + 1.7 + 4.00141= 17.72141

7.71 x 4.31 x 1.7 x 4.00141=  226.04433255

Answer:

FCC.

Explanation:

Hello,

In this case, since the density is defined as:

[tex]\rho =\frac{n*M}{Vc*N_A}[/tex]

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

[tex]Vc_{BCC}=(\frac{4r}{\sqrt{3} } )^3=(\frac{4*0.1345x10^{-7}cm}{\sqrt{3} } )^3=2.997x10^{-23}cm^3\\\\Vc_{FCC}=(2\sqrt{2}r)^{3} =(2\sqrt{2} *0.1345x10^{-7}cm)^3=5.506x10^{-23}cm^3[/tex]

Hence, we compute the density for each crystal structure:

[tex]\rho _{BCC}=\frac{n_{BCC}*M}{Vc_{BCC}*N_A}=\frac{2*102.9g/mol}{2.337x10^{-23}cm^3*6.022x10^{23}/mol} =14.62g/cm^3\\\\\rho _{FCC}=\frac{n_{FCC}*M}{Vc_{FCC}*N_A}=\frac{4*102.9g/mol}{5.506x10^{-23}cm^3*6.022x10^{23}/mol} =12.41g/cm^3[/tex]

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.

Answer:

[H₂SO₄] = 4.56 M

Explanation:

35.2 % by mass of H₂SO₄ means that in 100 g of solution, we have 35.2 grams of solute.

We convert the mass of solute to moles.

35.2 g / 98 g/mol = 0.359 moles

These moles are contained in 100 g of solution, so we use density to determine the volume.

1.27 g/mL = 100 g / volume

Volume = 100 g / 1.27 g/mL = 78.74 mL

Molarity is mmol /mL (either we can say, moles in 1L of solution).

We convert the moles to mmoles → 0.359 mol . 1000 = 359 mmoles

M = 359 mmoles/74.74 mL = 4.56 M

Answer:

23.28 g of O2.

Explanation:

We'll begin by calculating the mass of hexane. This can obtain as follow:

Volume of hexane = 10 mL

Density of hexane = 0.66 g/mL

Mass of hexane =?

Density = mass /volume

0.66 = mass of hexane /10

Cross multiply

Mass of hexane = 0.66 x 10

Mass of hexane = 6.6 g

Next, we shall write the balanced equation for the reaction. This is given below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:

Molar mass of C6H14 = (12.01x6) + (1.008 x 14)

= 72.06 + 14.112

= 86.172 g/mol

Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608 g

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.

Therefore, 23.28 g of O2 is needed for the reaction.

Answer:

107.55 J, 88.67 cal

Explanation:

1 Joule = 0.239 Calories

450 cal x 0.239 = 107.55 J

3.5 x 106 = 371

371 J = 371 x 0.239 cal

371 J = 88.67 cal

Answer:

B. S° > 0, ?H° < 0.

Explanation:

Expression for change in free energy is

     ΔG = ΔH - TΔS

For a reaction to be spontaneous , ΔG should be negative . When we watch the relation above , we find that for  ΔG will be negative at any temperature , if ΔH is negative and  ΔS is positive . Then both the terms on the right hand side will be negative and then   ΔG will become negative.

So option B is correct .

Answer:

Explanation:

Given that for Part A,

the arrangement can be either AgF or NaF.

On the off chance that the arrangement is AgF ,

For this situation AgF conc. can be controlled by titration with standard choride arrangement with potassium chromate as marker.

Get ready 1 M NaCl standard arrangement by dissolving 58.5 gm of NaCl in one liter refined water. Take 25 mL of 1 N NaCl arrangement in a funnel shaped carafe include hardly any drops of potassium chromate marker.

Take the example in the burette and titrate until a perpetual ruddy earthy colored hasten is seen.

[AgF responds with NaCl as follows

AgF + NaCl \small \rightarrow AgCl ( white precipitate)+ NaF

For whatever length of time that CaCl is available in the arrangement Ag structures AgCl hasten as above. When all the chloride particles expelled this way, the overabundance silver structures Ag2CrO4 ( rosy earthy colored encourage ) with the marker.

The dissolvability of AgCl is a lot of lower than the solvency of Ag2CrO4. Consequently the later won't hasten until all chloride particles exhausted.]

Let V2 be the volume of test utilized. At that point grouping of AgF in the example is given by

C2 = V1C1/V2 = 25 x 1/V2.

Rehash the titration by making legitimate weakening if any required.

On the off chance that the arrangement is NaF (Indirect strategy)

Here the Fluoride particle in a referred to volume of test is hastened as leadchlorofluoride with NaCl and PbNO3.

NaF + NaCl + Pb(NO3)2 \small \rightarrow PbClF (hasten) + 2 NaNO3

The above hasten is separated and washed and the washed encourage is taken in a measuring utencil and re-broke up in nitric corrosive to deliver the chloride particles.

PbClF + 2 HNO3 \small \rightarrow Pb(NO3)2 + HCl + HF

Presently include a known abundance of Silver nitrate answer for the above blend to encourage the chloride as silver chloride. Channel wash and evacuate the AgCl encourage.. The Filtrate contains the overabundance silver nitrate is to be dictated by titration with Standard chloride arrangement as in the past technique.

From the first amount of silver nitrate taken and the overabundance silver nitrate decided from the titration, we can decide the amount of silver nitrate responded.

The moles of silver nitrate responded = the moles of Cl delivered = the moles of F in the example.

In this way the convergence of sodium fluoride in the obscure example might be resolved.

Answer:

1219.5 kj/mol

Explanation:

To reach this result, you must use the formula:

ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).

The BE values are:

BE C = C: 839 kj / mol

BE C-H: 413 Kj / mol

BE O = O: 495 kj / mol

BE C = O = 799 Kj / mol

BE O-H = 463 kj / mol

Now you must replace the values in the above equation, the result of which will be:

ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol

Based on the bond energies, the reaction

HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of - 1222 kJ/mol.

The bond energies (E) of the reactions can help to calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.

[tex]\triangle H_ \degrees {rxn} = \sum E_{reactants} - \sum E_{products}[/tex]  

Solution:

bond energies in the formula:

[tex]\triangle H_ \degrees {rxn} = 835+ 2(411) + 2.5(494) - 4(799) - 2(459)[/tex]

= 1222 kJ/mol  

Thus, Based on the bond energies, the reaction:

HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of -1222 kJ/mol.  

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Answer:

1549.8 Joules.

Explanation

Energy transferred =  mass ( in kg) * specific heat of water * change in temperature in ( celsius).

= 0.03 * 4200 * (25 - 12.7)

= 1549.8 Joules.

According to specific heat capacity, 1549.8 joules of  energy is transferred when 30.0 g of water is cooled from 25.0°C to 12.7°C.

Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.

It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.

It is given by the formula ,

Q=mcΔT, substitution in formula gives, Q=30×4.2×12.3=1549.8 joules.

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Answer:

yes it is as it has a chemical formula which cannot be changed

Explanation:

Chromatography = It is a method in which a drop of mixture is spot is spotted at one of the end of the paper. After some time whenever it gets dried, it is then dipped into solvent without toushing the spot. As a result the mixtures gets separated.

Hence, the correct option is (c) "The separation of substances in a mixture by differences in their attraction to a substance over which they are passed"

Answer: Oxygen gas

Explanation: Gradpoint

Answer:

THE SPECIFIC HEAT OF THE METAL IS 28.69 J/g C

Explanation:

Mass of unknown metal = 30 g

Mass of water = 100 g

Initial Temperature of metal = 110 C

Initial temperature of water  = 200 C

Final temperature of the system = 25 C

Specific heat of water = 4200 J/ g C

Specific heat of metal = unknown

Heat lost = Heat gained\

M C Q = M C Q

30 * C * ( 25 - 110) = 100 * 4200 * (25 -200)

C = 100 * 4.18 * -175 / 30 * - 85

(the minus sign cancels out and we have):

C = 100 * 4.18 * 175 / 30 * 85

C = 73150 / 2550

C = 28.69 J/ g C

The specific heat of the unknown metal is hence 28.69 J/ g C

Answer:

ΔH = ΔH₁ + ΔH₂ - ΔH₃

Explanation:

Given that:

1. A → 2B

2. B → C + D

3. E → 2D

Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.

To estimate the ΔH for the process A → 2C + E

We multiply 2 with equation 2 where (B → C + D)

2B → 2C + 2D ⇒ 2ΔH₂

Also, let's switch equation (3), such that we have,

2D → E -ΔH₃

The summation of all the equation result into :

A → 2C + E

where; ΔH = ΔH₁ + ΔH₂ - ΔH₃

Based on Hess law of constant heat summation, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3

According to Hess' law of constant heat summation, the heat of any reaction ΔH∘f for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.

Thus, ΔH for the process A → 2C + E is calculated as follows:

The A → 2C + E can be obtained from the summation of the processes above:

A ---> 2B

2B ---> 2C + 2D

2D ---> E by reversing reaction step 3

Thus, ΔH = ΔH1 + ΔH2 - ΔH3

Therefore, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3

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Explanation:

1 mile = 1.61 km and 1 km = 1000 m

The speed of light is [tex]3\times 10^8\ m/s[/tex]. We need to convert it into miles per hour.

1 m = (1/1000) km

[tex]3\times 10^8\ m=\dfrac{3\times 10^8}{1000}\\\\=3\times 10^5\ km[/tex]

1 km = (1/1.61) miles

[tex]3\times 10^5\ km =\dfrac{3\times 10^5}{1.61}\\\\=1.86\times 10^5\ \text{miles}[/tex]

So, [tex]v=1.86\times 10^5\ \text{miles/s}[/tex]

Hence, this is the required solution.