The density of a nickel coin was determined by laboratory measurement to be 728 g/mL. A nickel has a mass of 5.040 grams. What is the volume of the nickel?
0.6923 mL
0.692 ml
оооо
36.3 mL
1.43 mL

Answer: • Phe can undergo oxidation to form Tyr.

• The Ala side chain does not form hydrogen bonds with other amino acids.

• Thr and Ser are polar amino acids.

• Lysine has one stereocenter (chiral center).

Explanation:

Amino acids are simply organic compounds which are known to contain carboxyl and amine functional groups and the main elements that are found in an amino acid are oxygen, carbon, hydrogen, and nitrogen.

The true statements about amino acids are:

• Phe can undergo oxidation to form Tyr.

• The Ala side chain does not form hydrogen bonds with other amino acids.

• Thr and Ser are polar amino acids.

• Lysine has one stereocenter (chiral center).

It should be noted that Methionine is not a thiol but rather a thio ether. Also, proline is simply a neutral amino acid and the statement that proline has an overall charge at physiological

pH(7.4) is false.

Answer:

XY₃

Explanation:

We have cations of the X³⁺ type and anions of the Y⁻ type. For a compound to be formed, it has to be electrically neutral, that is, it must have the same number of positive charges and negative charges so that the final charge of the compound is zero. In this case, the compound would be neutral if it had 1 atom of X³⁺ and 3 atoms of Y⁻. The resulting chemical formula would be XY₃.

Answer:

Explanation:

The density of a substance can be found by using the formula

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

Making mass the subject we have

From the question

Density = 1.98 g/mL

volume = 265 mL

Substitute the values into the above formula and solve for the mass

That's

mass = 1.98 × 265

We have the final answer as

Hope this helps you

Answer:

0.719M AgNO₃

Explanation:

The balanced chemical equation is:

MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂

Where 1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃

Molarity is the ratio between moles of solution and its volume in liters (The volume of the solution is 13.9mL = 0.0139L). That means all we need is to determine moles of silver nitrate. It is possible to know by using the chemical equation, thus:

Moles AgNO₃:

Moles of MgBr₂:

50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.

As 2 moles of AgNO₃ reacts per mole of  MgBr₂ and the reaction occurs completely, moles of silver nitrate are:

0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =

And molarity is:  

0.0100 moles AgNO₃ / 0.0139L =

Answer:

[tex]3.427 \times {10}^{3} [/tex]

Answer:

ΔH = 125.94kJ

Explanation:

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:

1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ

2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ

-1/2 (1):

WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ

3/2 (2):

3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ

The sum of  last both reactions:

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

ΔH = 842.7kJ -716.76kJ

The enthalpy change (ΔH) for the reaction is 125.94kJ.

Hess's Law:

The Hess's law states that the total enthalpy change during a complete chemical reaction is the same regardless of the path taken by the chemical reaction.

It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law).

It is given that:

1. 2W(s) + 3O₂(g) → 2WO₃(s) ΔH = -1685.4 kJ

2. 2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -477.84 kJ

On dividing equation 1. by - 1/2 for simplification we get:

W(s) + 3/2 O₂(g) → WO₃(s)  ΔH = 842.7kJ....................(a)

On dividing equation 2. by  3/2 for simplification we get:

3H₂(g) + 3/2O₂(g) → 3H₂O(g) ΔH = -716.76kJ...................(b)

On adding equations (a) and (b)

WO₃(s) + 3H₂(g) → W(s) + 3H₂O(g)

ΔH = 842.7kJ -716.76kJ

ΔH = 125.94kJ

The enthalpy change for the reaction is 125.94 kJ.

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Answer:

395.84cm^3

Explanation:

volume of a cylinder:

V=π(r^2)h

V = π((6cm)^2)(3.50cm) = 395.84cm^3

Answer:

second order

Explanation:

units of reaction and their order.

Zero order --> M^1 s^-1 = M/s

First order --> M^0 s^-1 = 1/s

Second order --> M^-1 s^-1 = L/mol s

In the question rate constant k =  4.65 L mol-1s-1. = 4.65 L/mol s

Hence, the reaction is a second order reaction

Answer:

The answer is "second".

Explanation:

Given:

rate constant for a reaction = 4.65 [tex]\frac{L}{mol \cdot s}[/tex]

[tex]\text{The rate of unit} = \frac{m}{s} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{where} \ \ M=\frac{mol}{l} \\\\[/tex]

                          [tex]=\frac{mol}{l \cdot s}[/tex]

The orders                                                       unit value of the k

0                                                                               [tex]\frac{m}{s}[/tex]

1                                                                                 [tex]\frac{1}{s}[/tex]

2                                                                               [tex]\frac{l}{m s}[/tex]

The rate of the unit is constant for the second-order that's why it is correct.

Answer:0.3348%

Explanation:

Answer:

Mass, m = 37900 grams

Explanation:

Given that,

The density of a brand of gasoline is 0.737 g/mL at 25°C

We need to find mass of gasoline would fill a 13.6 gal tank.

Density is defined as mass per unit volume.

We know that, 1US gal=3.78L

So,

1US gal = 3785.41 mL

13.6 US gal = 51481.6 mL

So,

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\m=0.737\ g/mL\times 51481.6\ mL\\\\m=37941.93\ g[/tex]

or

m = 37900 g (in three significant figures)

So, the mass of the gasoline is 37900 grams.

Mass, m = 37900 grams

Given that,

The density of a brand of gasoline is 0.737 g/mL at 25°C

 find the mass of gasoline would fill a 13.6 gal tank.

Density is defined as mass per unit volume.

We know that, 1US gal=3.78L

So,

1US gal = 3785.41 mL

13.6 US gal = 51481.6 mL

So,

[tex]Density=\frac{mass}{volume}\\\\0.737\ g/mL=mass/51481.6\ mL\\mass=37941.9392\ g[/tex]

or

m = 37900 g (in three significant figures)

So, the mass of the gasoline is 37900 grams.

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Answer:

A- Perform an experiment

Explanation:

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

Answer:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid.

Answer:

In the mid-19th century Claude Bernard was also influential, especially in bringing the scientific method to medicine. In his discourse on scientific method, An Introduction to the Study of Experimental Medicine (1865), he described what makes a scientific theory good and what makes a scientist a true discoverer.

Explanation:

Answer:

If the pressure in the tire is low, then the tire lost air pressure because air is escaping through a hole in the tire.

Explanation:

The tire will lose air from a puncture in the tire, rather than the the second hypothesis saying tires do not lose air naturally.

Answer:

1) If the pressure in the tire is low, then the tire lost air pressure because air is escaping through a hole in the tire

Explanation:

I JUST DID IT

Answer:

1.51 × 10⁻⁵

Explanation:

Step 1: Given data

pKa value: 4.82

Step 2: Write the reaction for the acid ionization of butanoic acid, HC₄H₇O₂

Butanoic acid is a weak acid that ionizes according to the following equation.

HC₄H₇O₂(aq) ⇄ C₄H₇O₂⁻(aq) + H⁺(aq)

The acid ionization constant for this reaction is Ka

Step 3: Calculate the value of the ionization constant for butanoic acid

We will use the following expression.

pKa = -log Ka

Ka = antilog -pKa = antilog -4.82 = 1.51 × 10⁻⁵

Answer:

pH = 4.58

Explanation:

The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

Moles that react:

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles

CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

Where [] is molarity of each species or moles

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

Answer:

See attached picture.

Explanation:

Hello,

In this case, on the attached picture, you will find the required structure wherein we can see the ethynyl radical at the first carbon which is composed by two carbon atoms bonded via a triple bond due to the "ynyl" suffix. Moreover, the cyclohexane acts as the main chain.

Best.

Answer:

324.46 amu

3.33 × 10⁻³ mol

Explanation:

(a) Calculate the molecular mass of quinine, C₂₀H₂₄N₂O₂

We will add the mass of each atom times the number of each atom in a molecule.

20 × C + 24 × H + 2 × N + 2 × O

20 × 12.01 amu + 24 × 1.01 amu + 2 × 14.01 amu + 2 × 16.00 amu

324.46 amu

(b) Calculate the number of moles of quinine in 1.08 g of quinine

The molar mass of quinine is 324.46 g/mol.

1.08 g × (1 mol/324.46 g) = 3.33 × 10⁻³ mol

Answer:

Carbon

Explanation:

It has 6 protons... carbon has 6 protons

Answer:

When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.

Answer:

Ethyl ethers are Non-polar solvent and can extract polar molecules such as fatty acids

Answer:

Explanation:

Density of a substance can be found by using the formula

From the question

mass = 5.52 g

volume = 12 mL

Substitute the values into the above formula and solve for the Density

That's

We have the final answer as

Hope this helps you

Answer:

Density is 0.46 g/mL

Explanation:

Answer:

62 lb

Explanation:

Step 1: Given data

Dosage of the intramuscular medication: 20 mg/kg of body weight

Mass of the intramuscular medication: 560 mg

Step 2: Calculate the body weight of the patient

560 mg IM × (1 kg BW/20 mg IM) = 28 kg BW

Step 3: Convert the body weight of the patient to pounds

We will use the relationship 1 kg = 2.20 lb.

28 kg × (2.20 lb/1 kg) = 62 lb

Answer:

The additional concentration of nitrogen in the river water due to the farmer's fertilizer is 0.0629 milligrams per liter

Explanation:

The mass of the fertilizer the farmer applies = 1705 kg

The percentage of nitrogen contained in the fertilizer = 10%

Therefore;

The mass of nitrogen contained in the fertilizer = 10% of 1705 kg = 10/100 × 1705 kg

The mass of nitrogen contained in the fertilizer = 10/100 × 1705 kg = 170.5 kg

The percentage of the fertilizer that washes into the river that runs through the farm = 15.0%

The mass of the fertilizer that washes into the river = 15/100 × 170.5 = 25.575 kg = 25575 g = 25575000 mg

The volume of water that flows through the farm = 0.455 cubic feet per second

The volume of water that flows through the farm in a year = 0.455 ft³ × 60 × 60 × 24 × 365 = 14348880 ft³ = 406,315.03 m³ = 406,315,034 l

The number of moles of nitrogen = Mass of nitrogen/(Molar mass of nitrogen)

Molar mass of nitrogen =14.0067 g/mol

Therefore;

The number of moles of nitrogen = 25575 kg/(14.0067 g/mol) = 1.825.912 moles

The additional concentration of the nitrogen in moles = 1.825.912 moles/(406,315,034 l) = 4.49 ×   10⁻⁶ mol/liter

The additional concentration of the nitrogen in milligrams of nitrogen per liter of water = 25575000/406,315,034 = 0.0629 milligrams of nitrogen per liter.

Therefore, the additional concentration of nitrogen in the river water due to the farmer's fertilizer = 0.0629 milligrams per liter

Answer:

The value is  [tex]K  =  8*10^{61}[/tex]

Explanation:

From the question we are told that

    The equation is  [tex]2 Cr  +  3Pb^{2+} \to 3Pb + 2Cr^{3+}[/tex]

     The  temperature is  [tex] T = 25^oC =  298 K   [room  \ temperature ][/tex]

     The  emf at standard condition is  [tex]E^o_{cell}  =  0.61 \  V[/tex]

Generally at the cathode

      [tex]3Pb^{2+}(aq) + 6 e- --> 3Pb(s)[/tex]

  At the anode

      [tex]2Cr^{3+} + 6e^- \to  2Cr[/tex]

Generally for an  electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as  

       [tex]G =  n*  F *  E^o_{cell} [/tex]

Here  n  is  the no of electron  with value n = 6

       F  is  the Faraday's constant with value 96487 J/V

  =>   [tex]G =  6  * 96487 *  0.61[/tex]

  =>   [tex]G = 3.5 *10^{5} \  J[/tex]

This Gibbs free energy can also be represented mathematically as

       [tex]G =  RTlogK[/tex]

Here  R  is the cell constant with value 8.314J/K

           K is the equilibrium constant

From above

=>  [tex]K  =  antilog^{\frac{G}{ RT} }[/tex]

Generally  antilog =  2.718

=>[tex]K  =  2.718^{\frac{3.5 *10^5}{ 8.314* 298} }[/tex]

=>   [tex]K  =  8*10^{61}[/tex]

Answer:

D. 13.44 dm³

Explanation:

Step 1: Write the balanced equation

CuO + H₂ ⇒ Cu + H₂O

Step 2: Calculate the mass corresponding to 47.7 g of CuO

The molar mass of CuO is 79.55 g/mol.

47.7 g × (1 mol/79.55 g) = 0.600 mol

Step 3: Calculate the moles of H₂ that react with 0.600 mol of CuO

The molar ratio of CuO to H₂ is 1:1. The moles of H₂ are 1/1 × 0.600 mol = 0.600 mol.

Step 4: Calculate the volume of 0.600 moles of H₂ at STP

1 mol of any gas at STP has a volume of 22.4 dm³.

0.600 mol × (22.4 dm³/1 mol) = 13.44 dm³

Answer:

a method of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.

Explanation:

a method of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.

Answer:

Both accuracy and precision

Explanation:

This displays accuracy because the weights were all very close to 12 oz, which was the desired weight/true value. The weights only varied from 12 oz by 0.1 or 0.2 ounces.

Additionally, this displays precision because the weights were very close to each other as well. The weights only varied slightly, by 0.1 to 0.3 ounces

answer:

Both accuracy and precision

Explanation:

This displays accuracy because the weights were all very close to 12 oz, which was the desired weight/true value. The weights only varied from 12 oz by 0.1 or 0.2 ounces.

Additionally, this displays precision because the weights were very close to each other as well. The weights only varied slightly, by 0.1 to 0.3 ounces

Answer:

75416 ft/min

(857 mi/h) x (5280 ft/h ÷ 1 mi / 1h ) x (1h x 1 ft ÷ 60 min x 1 ft)

Explanation:

Conversion of 857 mi/h to feet /second.

We'll begin by converting 857 mi/h to ft/h.

This is illustrated below:

Recall:

1 mi/h = 5280 ft/h

Therefore,

857 mi/h = 857 x 5280 = 4524960 ft/h

857 mi/h = 4524960 ft/h

Finally, we shall convert 4524960 ft/h to ft/min.

This can be obtained as follow:

Recall:

1 ft/h = 1/60 ft/min

Therefore,

4524960 ft/h = 4524960 x 1/60 = 75416 ft/min

4524960 ft/h = 75416 ft/min

Therefore, 857 mi/h is equivalent to 75416 ft/min.

To solve the above question in a shorter for we simply do the following:

(857 mi/h) x (5280 ft/h ÷ 1 mi/1h ) x (1h x 1 ft ÷ 60 min x 1 ft)

Answer:

0.09889

Explanation:

Order of Operations: BPEMDAS

Sig figs: 4

Step 1: Write out equation

[tex]\frac{3.434(1.245)}{43.234}[/tex]

Step 2: Multiply

[tex]\frac{4.27533}{43.234}[/tex]

Step 3: Divide

0.098888

Step 4: Round

0.09889