Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g)  NH3(g)
(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.

Answer:

39.0229 amu

Explanation:

Hello there!

In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:

[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]

Regards!

Answer:

(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms

Explanation:

Alcohols are poor leaving groups.

To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.

Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.

After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.

Answer:

Explanation:

First we convert the given masses into moles, using the compounds' respective molar mass:

Then we multiply each amount by Avogadro's number, to calculate the number of molecules:

Answer:

D

Explanation:

The longer handles distribute the force across a longer distance.

C. It can help us track how quickly element of the climate are changing

The equation for the energy required to melt 2 kg of gold is 3440 kJ.

Energy is the ability to do work or cause change. It is an essential part of everyday life and is present in many forms, such as thermal energy, electrical energy, chemical energy, and mechanical energy. Energy can be converted from one form to another in order to do work.

The equation for calculating the energy required to melt a certain mass of material is Q = m x Lf, where Q is the energy required (in joules), m is the mass of the material (in kilograms), and Lf is the latent heat of fusion (in joules per kilogram).
Using the table below, we can see that the latent heat of fusion for gold is 1760 kJ/kg. Therefore, the equation for the energy required to melt 2 kg of gold is: Q = 2 kg x 1760 kJ/kg = 3440 kJ.

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Answer:

0.518 g

Explanation:

Step 1: Write the balanced equation

3 CaCl₂ + 2 H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6 HCl

Step 2: Calculate the moles corresponding to 0.555 g of CaCl₂

The molar mass of CaCl₂ is 110.98 g/mol.

0.555 g × 1 mol/110.98 g = 5.00 × 10⁻³ mol

Step 3: Calculate the moles of Ca₃(PO₄)₂ produced

5.00 × 10⁻³ mol CaCl₂ × 1 mol Ca₃(PO₄)₂/3 mol CaCl₂ = 1.67 × 10⁻³ mol Ca₃(PO₄)₂

Step 4: Calculate the mass corresponding to 1.67 × 10⁻³ moles of Ca₃(PO₄)₂

The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol.

1.67 × 10⁻³ mol × 310.18 g/mol = 0.518 g

Answer:

a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.

Explanation:

It is given that a student performed an experiment where he dropped a nickel metal in to HCl solution. He observed the reaction and performed a splint test in the test tube that is filled with a gas which is formed while Nickle is dropped into the solution of HCl.

But the experiment that the student performed was incorrect. He must have observed the popping sound when the splint was placed in the test tube.

When the splint was added to the gas splint flared up. The hydrogen gas pops out when exposed to the flame.

[tex]$Ni + HCl(aq) = NiCl + H_2$[/tex]

Thus the correct option is (a).

Answer:

The pKa of an acid can be determined through titration with a strong base.

Gradually increase the volume of the base, stopping before the equivalence point is reached.

The pKa of the acid is equal to the pH at the midway volume to the equivalence point.

Explanation:

An acid HA dissociates in water as follows:

HA ⇄ H⁺ + A⁻      Ka

So, it produces hydrogen ions (H⁺) and a conjugate base (A⁻). The concentrations of HA, H⁺ and A⁻ at equilibrium determine the constant Ka. The pKa is calculated as:

pKa = -log Ka

The relationship between the pH of the solution and the pKa of the acid is described by the Henderson-Hasselbalch equation:

pH = pKa + log ([A⁻]/[HA])

The pKa can be experimentally determined by acid-base titration, in which a strong base is added to the acid solution. As the base is added, the acid HA is neutralized and the conjugate base A⁻ is formed. Thus, the concentration of the acid ([HA]) increases and the concentration of the conjugate base ([A⁻] decreases. The equivalence point is reached when the total amount of acid is neutralized with the added base. Before reaching the equivalence point, at the halfway point, half of the acid is neutralized and converted into the conjugate base. Thus:

[A-] = [HA] ⇒ log [A-]/[HA] = log 1 = 0 ⇒ pH = pKa

We measure the pH at that point and it is equal to the pKa of the acid.

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

We have to find number of moles of O2 needed  to produce 12 moles of H2O.

From given equation

We can see that

6 moles of   H2O produced by Oxygen =7 moles

1 mole of   H2O produced by Oxygen=[tex]\frac{7}{6}[/tex]moles

12 moles of H2O produced by Oxygen=[tex]\frac{7}{6}\times 12[/tex]moles

12 moles of H2O produced by Oxygen=[tex]7\times 2[/tex]moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

The amount of oxygen required for the combustion of ethane to produce 12 moles of water is 14 moles.

The moles of oxygen produced in the reaction can be given from the stoichiometric law of the balanced chemical equation.

The balanced chemical equation for the combustion of ethane is:

[tex]\rm 2\;C_2H_6\;+\;7\;O_2\;\rightarrow\;4\;CO_2\;+\;6\;H_2O[/tex]

The 6 moles of water are produced from 7 moles of oxygen. The moles of oxygen required to produce 12 moles of water are:

[tex]\rm 6\;mol\;H_2O=7\;mol\;Oxygen\\12\;mol\;H_2O=\dfrac{7}{6}\;\times\;12\;mol\;O_2\\ 12\;mol\;H_2O=14\;mol\;O_2[/tex]

The moles of oxygen required to produce 12 moles of water are 14 moles.

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Answer:

e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature

Answer:

D. Temperature and activation energy is the correct answer

Explanation:

^_^

Answer:

acidic titration in its stable form

Explanation:

Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.

Answer:

Throughout the explanations section below you will find a description of the question.

Explanation:

(1)

(2)

the answer is silicon!!

Answer:

7.6×10¹⁰

Explanation:

7.296×10²÷9.6×10⁻⁹

To solve such problem,

We group the whole number ans solved seperately and also group the indices and solve the seperately

Step1 : 7.296/9.6 = 0.76

Step 2: applying the law of indices,

10²÷10⁻⁹ = 10⁽²⁺⁹⁾ = 10¹¹

Therefore,

7.296×10²÷9.6×10⁻⁹ = 0.76×10¹¹ = 7.6×10¹⁰

Answer:

1. Its formation requires very strong updrafts = a. Hail

2. Its formation requires falling through a layer of above-freezing air = d. Freezing Rain

3. Precipitation from cumuliform clouds is typically of this nature = c. Shower

4. Precipitation from stratus clouds is typically of this nature = Drizzle

Explanation:

Hail formation requires very strong updrafts, these updrafts are the upward moving air created in a thunderstorm. This period of noticeable thunderstorms creates hails.

Freezing rain requires the presence of warm air, it requires falling through a layer of above-freezing air to the colder air below to produce an ice coating on anything it drops on.

Showers are produced by cumuliform clouds which look like cotton balls. Since cumuliform clouds precipitate too, these clouds can have fluctuating rain in a day in the form of showers.

Drizzle which raises low visibility is considered a type of liquid precipitation since it also falls from a cloud. Drizzle which is obviously smaller in diameter when compared to that of raindrops, however, is common with stratus clouds.

The question is incomplete, the complete question is shown in the image attached to this answer

Answer:

A

Explanation:

We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.

First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.

Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.

Thus, option A is the major product of the reaction.

Answer:

The hot water dissolves the flavor and color components.

Explanation:

Caffeine can be extracted from the tea leaves into hot water because the hot water dissolves the flavor and color components away from the solid vegetable. This is an example of a solid-liquid extraction. We can extracted the aqueous solution and isolated the crude caffeine by converting the components of caffeine into their calcium salts which are insoluble in water. Then the caffeine can be extracted from the water by using methylene chloride.

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

We can find the pH of the solution from the reaction of equilibrium (3).

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

We can find the pH of the solution from reaction (3).

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

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Answer:

The most important difference between the two is that butter is derived from dairy and is rich in saturated fats, whereas margarine is made from plant oils. ... If the margarine contains partially hydrogenated oils, it will contain trans fat, even if the label claims that it has 0 g.

Explanation:

(⌒_⌒;)

Answer:

0.203 is the mean of the concentration of the HCl solution

Explanation:

You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:

0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =

Answer:

The correct option is A

Explanation:

Some amino acids, called glucogenic amino acids, when catabolized convert there carbon backbones to tricarboxylic acid (TCA) cycle intermediates. These intermediates can be subsequently metabolized into carbon dioxide and water with the release of ATP or the formation of glucose (known as gluconeogenesis.

All amino acids (with the exception of leucine and lysine) are glucogenic and can thus generate the carbon backbones required for gluconeogenesis. Thus, the correct option is a.

Answer:

The volume of the gas is fixed and will not change.

Explanation:

The volume of the gas will not change because there is no change in temperature. Temperature increases the volume of gases enclosed in a container.

Answer:

104.8 g

Explanation:

From the question given above, the following data were obtained:

Initial temperature of copper (T꜀) = 275.1 °C

Mass of water (Mᵥᵥ) = 272 g

Initial temperature of water (Tᵥᵥ) = 21 °C

Equilibrium temperature (Tₑ) = 29.7 °C

Mass of copper (M꜀) =?

NOTE:

Specific heat capacity of copper (C꜀) = 0.385 J/gºC

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Finally, we shall determine the mass of the copper in the sample. This can be obtained as follow:

Heat loss by copper = Heat gained by water

M꜀C꜀(T꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

M꜀ × 0.385 (275.1 – 29.7) = 272 × 4.184(29.7 – 21)

M꜀ × 0.385 × 245.4 = 1138.048 × 8.7

M꜀ × 94.479 = 9901.0176

Divide both side by 94.479

M꜀ = 9901.0176 / 94.479

M꜀ = 104.8 g

Thus, the mass of the copper in the sample is 104.8 g

Answer:

See explanation

Explanation:

Equation of the reaction;

Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)

Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3

Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl

1 mole of Na2CO3 reacts with 2 moles of HCl

0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl

Hence Na2CO3 is the limiting reactant

Since there is 0.6 moles of HCl present, the number of moles of excess reagent=

0.6 moles - 0.4 moles = 0.2 moles of HCl

1 mole of Na2CO3 forms 1 mole of water

0.2 moles of Na2CO3 forms 0.2 moles of water

Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water

1 mole of Na2CO3 yields 1 mole of CO2

0.2 moles of Na2CO3 yields 0.2 moles of CO2

1 mole of CO2 occupies 22.4 L

0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP

Hence;

V1=4.48 L

T1 = 273 K

P1= 760 mmHg

T2 = 27°C + 273 = 300 K

P2 = 760 mmHg

V2 =

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 × 4.48 × 300/760 × 273

V2= 4.9 L

The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.

Answer:stow the hairspray

Explanation:

Answer:

B

Explanation:

Spontaneous in chemical reactions means without any external input.

Occurring without added energy. Hence, option B is correct.

A spontaneous reaction is a reaction that supports the formation of products under the conditions under which the reaction is happening.

Spontaneous Reaction- a reaction that favours the formation of products at the conditions under which the reaction is occurring.

A non-spontaneous reaction can be made spontaneous if it is inside a controlled environment, this is what happens in nuclear power plants that create atomic fusion and fission in chambers that are controlled to control different particles to create nuclear active rays.

Hence, option B is correct.

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Answer:

False

Explanation:

An object with a density greater than 1.00g/mL (greater than the density of water) will sink. An object with a density less than the density of water, will float.

If the water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the bottom, this statement is false.

The density of an actual content is its mass per unit volume. The most common symbol for density is d, but the Latin letter D can also be used.

Three of an object's most fundamental properties are mass, volume, and density. Mass describes how heavy something is, volume describes its size, and density is defined as mass divided by volume.

The density of something is a measure of how heavy it is in relation to its size. When an artifact is more dense than water, it plunges; when an object is less dense than water, it floats.

Density is a property of a substance that is independent of the amount of substance.

As in the given scenario, water is having density 1 g/mL and object in having density less then it so it will float on water.

Thus, the given statement is false as the material will not sink, rather it will float on water.

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Answer:

The correct answer is Option c (0.163 g).

Explanation:

Given:

Heat energy,

Q = 0.21 J

Specific heat,

c = 4.184 J/g°c

Change in temperature,

ΔT = 0.308°C

As we know,

⇒ [tex]Q=mc \Delta T[/tex]

By substituting the values, we get

 [tex]0.21=m\times 4.184\times 0.308[/tex]

     [tex]m=\frac{0.21}{0.308\times 4.184}[/tex]

         [tex]=\frac{0.21}{1.28867}[/tex]

         [tex]=0.163 \ g[/tex]