If an object has a mass of 243.8 g and occupies a volume of 0.125L, what will be the density of this object in gm/cm3?

Answer:

0.1 L

Explanation:

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

Concentration of stock solution (C₁) = 1.25 M

Volume of diluted solution (V₂) = 0.5 L

Concentration of diluted solution (C₂) = 0.25 M

Volume of stock solution needed solution (V₁) =?

The volume of the stock solution needed can be obtained as follow:

C₁V₁ = C₂V₂

1.25 × V₁ = 0.25 × 0.5

1.25 × V₁ = 0.125

Divide both side by 1.25

V₁ = 0.125 / 1.25

V₁ = 0.1 L

Therefore, the volume of the stock solution needed is 0.1 L

Answer:

a. minus 10

Explanation:

An element in group 13 = Boron ,valence electrons = 3 , therefore, valence electrons in group 13 = group no. -10

An element in group 18 = Neon, valence electrons = 8 , therefore, valence electrons in group 18 = group no. - 10

For groups 13 through 18, the number of valence electrons in an atom is equal to the group number minus 10. Therefore, option (A) is correct.

Valence electrons in an atom can be described as the electrons occupying the outer most electron shell of an atom while the electrons in the inner shell are core electrons. Lewis structures can be helpful to calculate the number of valence electrons.

Valence electrons can be filled in several electron shells as they are caused interaction between atoms and responsible for the formation of chemical bonds. Only valence electrons can contribute to the formation of a chemical bond and decide the reactivity of the element.

The general electronic configuration of group 13 is ns²np¹ has three valence electrons. It can be determined as group number - 10 = 13 - 10 = 3.

The general electronic configuration of group 18 is ns²np⁶ has eight valence electrons. It can be determined as group number - 10 = 18 - 10 = 8.

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it's 0.2cm different from the expected value, wich was 14.6cm

how many percent that are is calculated in the screenshot.

rounded it's 1.37%

Answer: The statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.

Explanation:

A chemical reaction in which heat energy  is released is called an exothermic reaction. For exothermic reactions, the value of [tex]\Delta H[/tex] is always negative.

A chemical reaction in which heat energy is absorbed is called an endothermic reaction. For endothermic reaction, the value of [tex]\Delta H[/tex] is always positive.

In endothermic reactions, energy required for breaking the bonds between reactants is less than the energy when products are formed due to which the value of [tex]\Delta H[/tex] remains positive.

Thus, we can conclude that the statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.

It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed. The correct option is B.

The above reaction is endothermic because more energy is produced when new bonds form in the products (H = 920 kJ/mol) than is required to break bonds in the reactants (H = -750 kJ/mol).

In an endothermic process, more energy than is generated during bond creation is absorbed from the environment to dissolve existing bonds. This causes a net absorption of energy, which cools the system.

The reaction takes more energy than it releases, proving its endothermic nature, as seen by the positive difference between the energy needed to dissolve bonds and the energy released during bond formation.

Thus, the correct option is B.

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Your question seems incomplete, the probable complete question is:

During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)

Group of answer choices

A. It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.

B. It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.

C. It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.

D. It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.

The question is incomplete, the complete question is;

What functional group results when cyclopentanone reacts with ethylamine in the presence of trace acid? A) cyanohydrin B) semicarbazone C) imine D) enamine E) oxime

Answer:

imine

Explanation:

An imine is an unsaturated amine. An imine contains the carbon- nitrogen double bond.

Imines are obtained when a carbonyl compound is condensed with NH3 or an amine. The reaction involves several steps in its mechanism.

Since cyclopentanone is a ketone (carbonyl compound) and ethylamine is an amine,in the presence of trace acid, condensation of the two compounds occur to yield an imine

Answer:

6.7855 * 10^18

Explanation:

There are 2 moles in H2

Avagadro's number: 6.022E23

8.14E42*1/6.022E23 * 1/2

=6.7855 * 10^18

Answer: The empirical formula of the compound becomes [tex]C_5H_{10}O_2[/tex]

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Let the mass of the compound be 100 g

Given values:

% of C = 58.8%

% of H = 9.87%

% of O = [100 - 58.8 - 9.87] = 31.33%

Mass of C = 58.8 g

Mass of H = 9.87 g

Mass of O = 31.33 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

To formulate the empirical formula, we need to follow some steps:

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol[/tex]

[tex]\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol[/tex]

[tex]\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol[/tex]

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles

[tex]\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5[/tex]

[tex]\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5[/tex]

[tex]\text{Mole fraction of O}=\frac{1.96}{1.96}=1[/tex]

Converting the mole fraction into whole numbers by multiplying them with 2.

[tex]\text{Mole fraction of C}=2.5\times 2=5[/tex]

[tex]\text{Mole fraction of H}=5\times 2=10[/tex]

[tex]\text{Mole fraction of O}=1\times 2=2[/tex]

The ratio of C : H : O = 5 : 10 : 2

Hence, the empirical formula of the compound becomes [tex]C_5H_{10}O_2[/tex]

The empirical formula of the substance is[tex]C_5H_{10}O_2[/tex].

Explanation:

Given:

A compound made up of 58.8% of carbon, 9.87% hydrogen, and oxygen.

To find:

The empirical formula of this substance.

Solution

The percentage of carbon in a given substance by mass = 58.8%

The percentage of hydrogen in a given substance by mass = 9.87%

The percentage of oxygen in a given substance by mass :

[tex]= 100\%-58.8\%- 9.87\%=31.33\%[/tex]

Consider 100 grams of a substance.

In 100 grams of substance:

The mass of carbon =58.8% of 100 g = 58.8 g

The mass of hydrogen = 9.87% of 100 g = 9.87 g

The mass of oxygen = 31.33% of 100 g= 31.33 g

The moles of carbon [tex]=\frac{58.8 g}{12.0107 g/mol}=4.90 mol[/tex]

The moles of hydrogen[tex]=\frac{9.87g}{1.00784g/mol}=9.79 mol[/tex]

The moles of oxygen[tex]=\frac{31.33g}{15.999g/mol}=1.96 mol[/tex]

let the empirical formula of the substance = [tex]C_xH_yO_z[/tex]

The value of subscript x:

[tex]=\frac{4.90 mol}{1.96 mol}=2.5[/tex]

The value of subscript y:

[tex]=\frac{9.79mol}{1.96 mol}=5[/tex]

The value of subscript z:

[tex]=\frac{1.96mol}{1.96 mol}=1[/tex]

The empirical formula of a substance :

[tex]C_{2.5}H_5O_1=C_{\frac{25}{10}}H_5O_1=C_{25}H_{50}O_{10}[/tex]

Reducing to the lowest whole numbers:

[tex]C_{25}H_{50}O_{10}=C_5H_{10}O_2[/tex]

The empirical formula of the substance is [tex]C_5H_{10}O_2[/tex].

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Answer: The property of all bases is that they are substances which neutralizes an acid to form a salt and water only.

Explanation:

A base is also s substance that can accept or combine with a proton; a proton acceptor.

Bases include the oxides, hydroxides and carbonates of metals. These include MgO,Na2O, NaOH,Ca(OH)2. Most metals burn in oxygen to form metallic oxides which are basic. Examples are the MgO and Na2O.

Other properties or characteristics of bases includes:

--> They have a bitter taste.

--> They turn red litmus paper blue.

--> They are soapy to touch.

--> Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.

Bases (sodium hydroxide) can be used in the preparation of soaps, glass, paper and rayon. While some bases (magnesium hydroxide) can also be used to manufacture toothpaste and laxatives.

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:

Te

K

Explanation:

Main group elements are elements designated and located on the s-block and p-block on the periodic table. The core group or prominent elements are the elements that make up those groups. These groupings comprise the most naturally rich and abundant elements. The main group elements are those whose characteristics are more predicted as a function of their periodic table location. From the given options, only Te and K are elements of the main-group.

Answer:

answer number C is the correct answer for this

Answer:

The chemical formula of the unknown ionic compound is Ag₂CO₃.

Explanation:

The method used to identify the ions present in a given solution of an unknown compound is known as qualitative analysis. Specific tests are carried based on the chemical reaction of these ions and the results they give in these reactions, the ions present innthe compound are identified. Some of the results or observations include formation of precipitates on addition of certain reagents, evolution of gases, colour changes, as well as the color changes that are observed in a flame test of a solutionmof these compounds.

In the given question, when testing for cations, a white precipitate is formed when HCl is added to the unknown compound. When the white precipitate is heated in hot water, it does not dissolve.

From the table of observations, both Ag+ and Pb²+ give a white precipitate on addition of HCl but Zn²+ does not. So the cation present is either Ag+ or Pb²+. However, when heated in hot water, the white precipitate does not dissolve as will be observed if Ag+ were the cation present. Thus, the cation is Ag+.

When testing for anions, a white precipitate is formed when Ca(NO3)2 is added to the unknown compound and when HNO3 is added to the precipitate, it dissolves and bubbles are formed.

Now comparing these results with that as observed with CO₃²- and PO₄³-, both form white precipitates on addition of Ca(NO₃)₂, and the precipitate dissolves on addition of HNO₃. However, only CO₃²- dissolves with the production of bubbles when HNO₃ is added. Thus, the anion is CO₃²-.

Therefore, the compound is Ag₂CO₃ which is formed from the cation Ag+ and CO₃²-.

Answer:

3.65 g of cyclohexene

Explanation:

Cyclohexanol + phosphoric acid ----> cyclohexene

The reaction is 1:1 hence the limiting reactant is phosphoric acid.

Hence,

1 mole of phosphoric acid yields 1 mole of cyclohexene

0.0444 moles of phosphoric acid yields 0.0444 moles of cyclohexene

Theoretical yield = number of moles of cyclohexene × molar mass of cyclohexene

Theoretical yield = 0.0444 moles of cyclohexene × 82.143 g/mol

Theoretical yield = 3.65 g of cyclohexene

Answer:

1. not enough dye was added to the drink.

The wrong dye was added to the drink

the water in the drink is evaporating

2. Changing the compound changes the absorbance behavior.

3. Measure the absorbance for the same solution in different cuvette sizes and find the y-intercept.

Explanation:

When the beverage company adds dye to the drink, there should be standard quantity added to the drink so that the color of the drink remains constant. When too much dye is added to the drink, the color will get dark brown or black. When the color of drink get lighter than green this means dye is not added in required quantity.

Answer:

34.03 g

Explanation:

We'll begin by converting 635 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

635 mL = 635 mL × 1 L / 1000 mL

635 mL = 0.635 L

Thus, 635 mL is equivalent to 0.635 L

Next, we shall determine the number of mole of the solute (KBr) in the solution. This can be obtained as follow:

Volume = 0.635 L

Molarity = 0.450 M

Mole of KBr =?

Mole = Molarity × Volume

Mole of KBr = 0.450 × 0.635

Mole of KBr = 0.286 mole

Finally, we shall determine the mass of 0.286 mole of KBr. This can be obtained as follow:

Mole of KBr = 0.286 mole

Molar mass of KBr = 39 + 80

= 119 g/mol

Mass of KBr =?

Mass = mole × molar mass

Mass of KBr = 0.286 × 119

Mass of KBr = 34.03 g

Thus, the mass of the solute (KBr) in the solution is 34.03 g

Answer: The balanced half-equations for silver + oxygen= silver oxide are:

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

Explanation:

The word equation is as follows.

silver + oxygen = silver oxide

In terms of chemical formulas this equation can be written as follows.

[tex]Ag + O_{2} \rightarrow Ag_{2}O[/tex]

The removal on electron(s) from an atom, ion or molecule in a chemical reaction is called oxidation.

The gain of electron(s) by an atom, ion or molecule in a chemical reaction is called reduction.

Hence, half-reaction equations for the given reaction is as follows.

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

As the number of atoms participating in the reaction are equal. Hence, the half-equations are balanced.

Thus, we can conclude that the balanced half-equations for silver + oxygen = silver oxide are:

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

Answer:

323.15 or just 323 depending on the teacher

Answer:

Organic fertilizers are those fertilizers that are not artificially produced but are natural and has carbon in them.

Fertilizers in the broad term are used to provide nutrient to the soil and boost growth of crops.

Some of the advantages of organic fertilizers to inorganic fertilizers are:

They do not make crusts on the soil, unlike inorganic fertilizers.

They help add structure to the soil because of the good water movement into the soil

They are easier on the soil because they feed good microbes.

Answer:

0.350 L

Explanation:

First we convert 2.18 g of Na₂O₂ into moles, using its molar mass:

Then we convert 0.0279 moles of Na₂O₂ into moles of O₂, using the stoichiometric coefficients of the reaction:

Finally we use the PV=nRT formula to calculate the volume of oxygen produced:

Inputting the data:

Answer:

Earth has a mass of 5.9736×1024 kg

hope it is helpful to you

Answer:

vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.

Answer:

The molar mass of N2O5 is 108 g/mol. 1296g of N2O5 has 1296 / 108 = 12 moles. 1 mole contains 6.022 x 10^23 molecules

Answer:

LICK

Explanation:

Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

Explanation:

Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]

Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g

Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.

[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]

Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

Explanation:

Given:

Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]

To find:

The number of atoms of tantalum in 37.1 milligrams.

Solution:

Mass of tantalum = 37.1 mg

[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]

The number of atoms in 0.0371 grams of tantalum = N

Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]

Then a mass of N atoms of tantalum will be:

[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

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Explanation:

hope the picture above helps you to understand:)

Answer:

Explanation:

The mechanism of the reaction is shown in the diagram below. From the reaction, when (2S,3S)-2-Bromo-3-phenylbutane undergoes a reaction with sodium ethoxide (ETONa), the E2 elimination reaction is put into place. Here, the H and the leaving group are antiperiplanar to one another and the reaction mechanism proceeds to form an isomeric (E)-2-phenyl-2butane as the major product.