The southernmost point of india is _____
(a) indra point
( b) Gulf
( c) Amaravati
(d) jarawas​

0Answer: 0.6094

Explanation:

no of moles = mass / molar mass = 100/164.088= 0.6094 mole

Answer:

0.609 moles

Explanation:

mass in g ÷ atomic mass = moles

Ca(NO₃)₂ = 1 Ca 40.078 amu

2 N 28.0134 amu

+ 6 O 95.994 amu

____________________

164.0854 amu

100 g ÷ 164.0854 amu = 0.609 moles

three significant digits

Answer:

An atom of an element that exhibits typical metallic properties, being typically shiny, with high electrical and thermal conductivity. ... A metal may be a chemical element such as iron; an alloy such as stainless steel; or a molecular compound such as polymeric sulfur nitride.

Answer:

18,9g de hexano son necesarios

Explanation:

Basados en la ley de Raoult, la presión de vapor ejercida por una solución es:

P = P°*Xsolvente

Donde P es la presión de la solución deseada = 700mmHg, P° la presión de vapor de vapor del benceno = 1atm = 760mmHg y X es la fracción molar del solvente (Benceno).

Reemplazando:

700mm Hg = 760mmHg * X(Benceno)

0.9211 = X(Benceno)

La fracción molar de benceno se define como:

X = Moles benceno / Moles benceno + Moles hexano

Moles benceno -Masa molar: 78g/mol-

200g * (1mol/78g) = 2.5641 moles benceno

X = Moles benceno / Moles benceno + Moles hexano

0.9211 = 2.5641 moles benceno / 2.5641 moles benceno + Moles hexano

0.9211moles Hexano + 2.3618 = 2.5641

0.9211*Moles Hexano = 0.2023

Moles hexano = 0.2023/0.9211 = 0.2196 moles hexano.

Masa Hexano -Masa molar: 86g/mol-

0.2196 moles hexano * (86g/mol) =

La cantidad de hexano (C6H14) que se debe agregar a 200 g de benceno para permitir que la presión de vapor se convierta en 700 mm de Hg sería:

18.9 g

Usando la ley de Raoult, la presión de vapor que libera una solución se encuentra por:

P = P ° × X disolvente  

Donde (P) denota la presión de la solución deseada

P ° denota la presión de vapor del benceno y

X denota la fracción molar del disolvente (benceno).

En el caso dado,

(P) = 700 mmHg,

P° = 1atm = 760mmHg

Fracción molar de benceno = 0,9211           (∵ 700/760)

La fracción molar de benceno (X)  = Moles de benceno/Moles de benceno + Moles de hexano

Moles de benceno - Masa molar: 78g / mol -  200 g × (1 mol / 78 g)

= 2.5641 moles de benceno

X = Moles de benceno/Moles de benceno + Moles de hexano

0.9211 = 2.5641 moles Benceno/2.5641 moles Benceno + Moles de Hexano

⇒ 0,9211 × moles de Hexano + 2,3618 = 2,5641    

⇒ 0.9211 × moles de Hexano = 0.2023

∵ Moles de Hexano = 0.2023/0.9211

[tex]= 0.2196 moles[/tex]

[tex]Hexane mass - Molar mass = 86g/mol - 0.2196 mole hexane[/tex] × [tex](86g/mole)[/tex]

[tex]= 18.9g[/tex]

Por tanto, 18,9 g es la respuesta correcta.

Learn more about "Benzene" here:

brainly.com/question/25798187

Answer:

Synthetic sponges are made of three basic ingredients: cellulose derived from wood pulp, sodium sulphate, and hemp fiber. Other materials needed are chemical softeners, which break the cellulose down into the proper consistency, bleach, and dye.

Explanation:

IDK, BUT HOPE IT HELPS?

Answer:

The pyruvate dehydrogenase kinase enzyme activity would increase, resulting in an inhibition of pyruvate dehydrogenase activity. ... An in vitro study shows that isocitrate dehydrogenase is activated in the citrate cycle.

Explanation:

Answer:

1.78 mol

Explanation:

Step 1: Write the balanced equation

CaC₂ + H₂O ⇒ C₂H₂ + CaO

Step 2: Calculate the moles corresponding to 46.3 g of C₂H₂

The molar mass of C₂H₂ is 26.04 g/mol.

46.3 g × 1 mol/26.04 g = 1.78 mol

Step 3: Calculate the moles of H₂O required to form 1.78 moles of C₂H₂.

The molar ratio of H₂O to C₂H₂ is 1:1. The moles of H₂O required are 1/1 × 1.78 mol = 1.78 mol.

Answer:

In nomenclature of simple molecular compounds, the more electropositive atom is written first and the more electronegative element is written last with an -ide suffix.

The Greek prefixes are used to dictate the number of a given element present in a molecular compound.

Prefixes can be shortened when the ending vowel of the prefix “conflicts” with a starting vowel in the compound.

Common exceptions exist for naming molecular compounds, where trivial or common names are used instead of systematic names, such as ammonia (NH3) instead of nitrogen trihydride or water (H2O) instead of dihydrogen monooxide.

Terms

nomenclatureA set of rules used for forming the names or terms in a particular field of arts or sciences.

electronegativeTending to attract electrons within a chemical bond.

electropositiveTending to not attract electrons (repel) within a chemical bond.

Chemical Nomenclature

The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning to what chemical compound the name refers. Each chemical name should refer to a single substance. Today, scientists often refer to chemicals by their common names: for example, water is not often called dihydrogen oxide. However, it is important to be able to recognize and name all chemicals in a standardized way. The most widely accepted format for nomenclature has been established by IUPAC.

Molecular compounds are made when two or more elements share electrons in a covalent bond to connect the elements. Typically, non-metals tend to share electrons, make covalent bonds, and thus, form molecular compounds.

Rules for Naming Molecular Compounds:

Remove the ending of the second element, and add “ide” just like in ionic compounds.

When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.

If there is only one of the first element, you can drop the prefix. For example, CO is carbon monoxide, not monocarbon monoxide.

If there are two vowels in a row that sound the same once the prefix is added (they “conflict”), the extra vowel on the end of the prefix is removed. For example, one oxygen would be monooxide, but instead it’s monoxide. The extra o is dropped.

Generally, the more electropositive atom is written first, followed by the more electronegative atom with an appropriate suffix. For example, H2O (water) can be called dihydrogen monoxide (though it’s not usually). Organic molecules (molecules made of C and H along with other elements) do not follow this rule.

Here are the prefixes in naming molecular compounds:

Mono- 1

Di- 2

Tri- 3

Tetra- 4

Penta- 5

Hexa- 6

Hepta- 7

Octa- 8

Nona- 9

Deca- 10

Molecular compounds are named using a systematic approach of prefixes to indicate the number of each element present in the compound.

I hope it helps ●~●

Answer:

i believe energy is your answer here

Answer:

Swimming pool is an example of solution

Explanation:

It has water (solvent) and dissolve ion (solute)

Answer:

it is a solution

Explanation:

the combining of (water and Cl and other pool chemicals) make it a solution.

Answer:

silicon and germanium.

Explanation :

The passage states that at extreme pressures an elemental solid assumes the structure and bonding characteristics of a heavier element in the same column of the periodic table. The structure and bonding of diamond, which is a form of carbon, would therefore be most like other elements in the same group of the periodic table as carbon. This group contains silicon and germanium.

Answer:

Longer

Explanation:

:)

Answer:

Explanation:

Because you only have one repeat unit, n=1. 2n-1 becomes 2(1)-1 which is equal to one, meaning one molecule of H2O is produced, as is shown by the top condensation polymerisation reaction.

If you had two repeat units, n=2 so 2n-1=3. Three H2O molecules are produced because you would need two molecules of each reactant so three condesation reactions would occur and three molecules of H2O would be released.

Answer:

Todas son correctas

Explanation:

I) Los compuestos del carbono son solubles en disolventes no polares como el hexano, benceno, éter, etc. VERDADERO. La mayoría de los compuestos de carbono son apolares, y basados en la regla: Similar disuelve similar, podemos presumir que la mayoría de compuestos de carbono se disuelven en solventes no polares.

II) Los compuestos del carbono generalmente presentan puntos de fusión y de ebullición bajos. VERDADERO. Al ser sustancias apolares, sus fuerzas electrostáticas son bajas. De la misma manera, como su masa atómica es pequeña, las fuerzas de Van der Waals son despreciables haciendo que sus puntos de fusión y ebullición sean bajos respecto a sustancias de estrucutra similar.

III) La gran mayoría de los compuestos del carbono son combustibles, sean estos, gaseosos, líquidos o sólidos. VERDADERO. Los gases (Como el gas natural) son combustibles usados para cocina. Los líquidos (Como la gasolina) son combustibles y bastante inflamables. Los sólidos (Como la madera) se usan como combustibles para hacer asados o son el combustible en incendios forestales.

IV) El enlace covalente es característico de los compuestos del carbono. VERDADERO. La polaridad del carbono es neutral haciendo que la mayoría de los enlaces que forma sean covalentes.

Answer:

Salts that produce acidic solutions are acid salts. Neutral salts are those salts that are neither acidic nor basic. Zwitterions contain an anionic and a cationic centre in the same molecule, but are not considered to be salts. Examples of zwitterions include amino acids, many metabolites, peptides, and proteins.

Explanation:

Answer:

Positive

Explanation:

Answer:

C. x⁴ + 6·x³ - 12·x  - 72

Explanation:

The given functions are;

[tex]f(x) =\sqrt{x^2 + 12 \cdot x + 36}[/tex]

g(x) = x³ -12

We have that [tex]f(x) =\sqrt{x^2 + 12 \cdot x + 36}[/tex] = [tex]f(x) =\sqrt{(x + 6)^2}[/tex] =  (x + 6)

Therefore;

f(x)·g(x) = [tex]\sqrt{x^2 + 12 \cdot x + 36}[/tex] × (x³ - 12) = (x + 6) × (x³ - 12)

(x + 6) × (x³ - 12) = x⁴ - 12·x + 6·x³ - 72 = x⁴ + 6·x³ - 12·x  - 72

∴ f(x)·g(x) = [tex]\sqrt{x^2 + 12 \cdot x + 36}[/tex] × (x³ - 12) = x⁴ + 6·x³ - 12·x  - 72

Answer:

highly concentrated solution of acid

Answer:its B

Explanation:

i have done this before! and i got 100 so i hope you got all the other questions right!

The table shows four reactions. The reaction That is an example of combustion of an organic substance is C₂H₄ + H₂ → C₂H₆. The correct option is b) 2.

Combustion reactions are those reactions in which the reactant reacts with oxygen and produces heat and light. An example is the burning of wood, it happens in the presence of oxygen.

Organic compounds are those compounds that are made by carbons. They are made by carbon-carbon or carbon-hydrogen bonds. They are naturally present in nature.

C₂H₄ + H₂ → C₂H₆

The equation above shows the reaction between methane and hydrogen, which is forming ethane in the presence of oxygen.

Thus, the correct option is b) 2. C₂H₄ + H₂ → C₂H₆.

To learn more about combustion reactions, refer to the link:

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[tex]\huge\fcolorbox{red}{pink}{Answer ♥}[/tex]

The carbon atom is unique among elements in its tendency to form extensive networks of covalent bonds not only with other elements but also with itself. ... Moreover, of all the elements in the second row, carbon has the maximum number of outer shell electrons (four) capable of forming covalent bonds.

Hope it helps uh ✌️✌️✌️

Gud mrng

Answer:

heat energy to keep you warm and light energy to be able to read your book

Explanation:

Answer:

0,005

Explanation:

number of moles=mass/molecular mass

therefore it's going to be 0,20g over 40g/mol

which is 0,20g/40g/mol... grams and grams will cancel making the answer 0,005moles

Answer:

5 kg of feather

Explanation:

Answer:

Equilibrium concentrations of the gases are

[tex]H_2S=0.596M[/tex]

[tex]H_2=0.004 M[/tex]

[tex]S_2=0.002 M[/tex]

Explanation:

We are given that  for the equilibrium

[tex]2H_2S\rightleftharpoons 2H_2(g)+S_2(g)[/tex]

[tex]k_c=9.0\times 10^{-8}[/tex]

Temperature, [tex]T=700^{\circ}C[/tex]

Initial concentration of

[tex]H_2S=0.30M[/tex]

[tex]H_2=0.30 M[/tex]

[tex]S_2=0.150 M[/tex]

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

[tex]H_2S=0.30+2x[/tex]

[tex]H_2=0.30-2x[/tex]

[tex]S_2=0.150-x[/tex]

At equilibrium

Equilibrium constant

[tex]K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]

Substitute the values

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

By solving we get

[tex]x\approx 0.148[/tex]

Now, equilibrium concentration  of gases

[tex]H_2S=0.30+2(0.148)=0.596M[/tex]

[tex]H_2=0.30-2(0.148)=0.004 M[/tex]

[tex]S_2=0.150-0.148=0.002 M[/tex]

Explanation:

2C +O2 = 2CO

this will be the answer

Answer:

C. T>617 K

Explanation:

We are given that

[tex]\Delta H=176KJ/mol[/tex]

[tex]\Delta S=0.285KJ/K\cdot mol[/tex]

We have to find the temperature at which the reaction is spontaneous.

When

[tex]\Delta H>0, \Delta S>0[/tex]

Therefore, the reaction is spontaneous at certain range of temperature.

Option A is not true.

[tex]\Delta G=\Delta H-T\Delta S[/tex]

When [tex]\Delta G[/tex] is negative, then  the reaction is spontaneous.

[tex]\Delta G=176-0.285T[/tex]

When T<50

Suppose T=49 K

[tex]\Delta G=176-49\times 0.285>0[/tex]

Therefore, [tex]\Delta G is positive[/tex].Hence, the  reaction is not spontaneous.

Option B is wrong.

C.T>617K

Suppose T=618 K

[tex]\Delta G=176-0.285\times 618=-0.13<0[/tex]

Therefore, [tex]\Delta G is negative[/tex].Hence, the  reaction is spontaneous.

So, option C is true.

D.T<617 K

Suppose T=616 K

[tex]\Delta G=176-0.285\times 616=0.44>0[/tex]

Therefore, [tex]\Delta G is positive[/tex].Hence, the  reaction is not spontaneous.

So, option D is not true.

Answer:

3 Protons, 2 Neutrons, & 3 Electrons. Is the charge of this atom positive, negative, or neutral?

Explanation:

Put these atoms in order from most negative overall charge to least negative

overall charge.

11 Atom X: 104 protons, 102 electrons

11 Atom B: 24 protons, 18 electrons

1 Atom Q: 15 protons, 16 electrons

11. Atom P: 7 protons, 10 electrons

Answer:

CH2O

Explanation:

According to this question:

C = 2.04g

H = 0.34g

O = 2.73g

First, we divide the mass value of each element by its atomic mass to convert to moles.

C = 2.04g ÷ 12 = 0.17mol

H = 0.34g ÷ 1 = 0.34mol

O = 2.73g ÷ 16 = 0.17mol

Next, we divide by the smallest mole value (0.17):

C = 0.17mol = 0.17 = 1

H = 0.34mol ÷ 0.17 = 2

O = 0.17mol ÷ 0.17 = 1

The whole number ratio of C,H,O is 1:2:1, hence, the empirical formula is CH2O.