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ANSWER:

(d) is the write answer

EXPLANATION:

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ANSWER:

here is your answer

water H2O

Answer:

approximately 40 to 45 mm Hg

PaCO2 is the carbon dioxide partial pressure in alveoli, which in NORMAL physiological conditions is approximately 40 to 45 mm Hg, and the RQ (respiratory quotient). FiO2 is directly related to the percent composition of oxygen in the inspired AIR. WITHOUT support at sea level, this is 21% or 0.21.

Explanation:

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Answer:

METALS below hydrogen (copper, silver, GOLD and PLATINUM), will not react with DILUTE acid. They cannot displace hydrogen from the non-metal ANION.

ANSWER: CH3COOH

Explanation: Vinegar CONTAINS acetic acid ( CH3COOH). When it is UNDILUTED, it is sometimes called glacial acetic acid.

Explanation:

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Answer:

The RELATIVE atomic mass in 1 MOLE of isotopes of sodium atoms is 23 g/mol.

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Answer:

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3KOH + H3PO4 → K3PO4 + 3H2O

ANSWER:

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ANSWER:

A. FE Pb Co Cu

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lyophillic and lyophobic colloids...

1. The AFFINITY between DISPERSE phase and dispersion medium is strong.

1. The affinity between disperse phase and dispersion medium is weak.

2. The viscosity of colloid is HIGHER than that of the dispersion medium.

2. The viscosity of colloid is same as that of the dispersion medium.

3. The surface tension of colloid is LOWER than that of the dispersion medium.

3. The surface tension of colloid is same as that of the dispersion medium.

Explanation:

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ANSWER:

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Answer:

OKKK

EXPLANATION:

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EXPLANATION:

the MIXTURE of FLOUR and WATER is the mixture of SOLID and liqued

ANSWER:

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Explanation:

The use of Plastic is a THREAT to environment because of the following reasons: ... Articles made up plastics such as plastic bags, bottles do not rot when they are thrown away after use. The waste plastic keep accumulating in the surroundings and pollute the environment . Thus, Plastics are not environment FRIENDLY

ANSWER:

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Answer:

An acid is a substance or COMPOUND that releases hydrogen IONS (H+) when in SOLUTION. In a strong acid, such as hydrochloric acid (HCL), all hydrogen ions (H+), and chloride ions (CL-) dissociate (separate) when placed in water and these ions are no longer held together by ionic bonding.

Answer:

THANKS for FREE point

Explanation:

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Answer:

The WORD ruby comes from ruber, Latin for red. The color of a ruby is due to the ELEMENT chromium. Some gemstones that are popularly or historically called rubies, such as the Black PRINCE's Ruby in the BRITISH Imperial State Crown, are actually spinels. These were once known as "Balas rubies".

EXPLANATION:

Hydride ion transfer

• The first partial reaction, the hydride transfer, requires the cofactor nicotinamide ADENINE dinucleotide PHOSPHATE (NADPH). NADPH serves as a source for two electrons and a PROTON (H(+)), the equivalent of a hydride ANION, H(-). Hydride transfer reactions play a key role in many biological systems.

ANSWER:

O

EXPLANATION:

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ANSWER:

These PEOPLE are very ILLUSTRATED

Answer:

Alternatively we can calculate pOH = - LOG [OH-]= - log [10-1] = 1 and remembering that pH + pOH = 14 we get pH = 14 - 1 = 13. FOLLOW me

<P>Answer:

The electronegativity order increases in the order:

• A. C, N, Si, P.
• A. C, N, Si, P.B. N, Si, C, P.
• A. C, N, Si, P.B. N, Si, C, P.C. Si, P, C, N.
• A. C, N, Si, P.B. N, Si, C, P.C. Si, P, C, N.D. P, Si, N, C.

Answer:

An ACID base reactions is a CHEMICAL reactions that occur between an acid and base . It can be used to be determine pH .

EXPLANATION:

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Answer:

Ratio of O2 : N2 by weight=1:4.

Let the weight of O2= 1 gm.

The weight of N2=4 gm.

The ratio of the No. Of MOLECULES will be the ratio of the Number of their moles ie.,

= 1/32 :4/28

= 28: (4 x32) = 7 : 32

= 7:3

Answer:

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11th

Chemistry

Equilibrium

Solubility Equilibria

In a mixture of N2 and H2 i...

CHEMISTRY

In a mixture of N

2

and H

2

initially in a mole ration of 1:3 at 30 atm and 300

o

C, the percentage of ammonia by volume under the equilibrium is 17.8. CALCULATE the equilibrium constant (K

P

) of the mixture, for the reaction, N

2

(g)+3H

2

(g)⇌2NH

3

(g)

HARD

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ANSWER

Let the initial moles of N

2

and H

2

be 1 and 3.

N

2

+3H

2

⟶2NH

3

Initial 1 3 0

at Eqb. 1−x 3−3x 2X

% of volume is same as % by mole NH

3

4−2x

2x

=0.178

x=

2+(2×0.178)

4×0.178

x=0.302

mole fraction of H

2

at eqb.=

4−2x

3−3x

=0.6165

Mole fraction of N

2

at eqb.=1−0.6165=0.178=0.2055

K

p

=(X

NH

3

.P

T

)

2

/(X

N

2

)(X

H

2

.P

T

)

3

=

(0.2055×30)(0.6165×30)

3

(0.178×30)

2

=7.31×10

−4

atm

−2

ANSWER:

Explanation:

12 G

ANSWER:

????????? I will THINK it LATER

Explanation:

we are developing new industries and manufacturing PLANTS to increase the goods PRODUCTIVITY. HARMFUL moderations: 1) development and expansion of the human locality is demolishing the natural VEGETATION. increasing industral plants around is also increasing the amount of pollution in our environment.

ANSWER:

yeshdjjfificicicucuc

yes

Answer:

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Explanation:

Nomenclature. The TRIVIAL name carbyne is the preferred IUPAC name. Following the substitutive nomenclature, the MOLECULE is viewed as methane with THREE hydrogen ATOMS removed, YIELDING the systematic name "methylidyne".

Answer:

Oxymercuration-demercuration: A reaction in which an ALKENE is reacted with a HG2+ salt and an oxygen nucleophile (water or an ALCOHOL), to form an ORGANOMERCURY INTERMEDIATE. The intermediate's carbon-mercury bond is then converted to a carbon-hydrogen bond by treatment with NaBH4.

ANSWER:

no dear scl 6 is know because CL has d ORBITAL bt F do not have d orbital..

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ANSWER:

OPTION 1

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Answer:

CONSIDER this redox REACTION:

Zn + Cu2+ → Zn2+ + Cu

If you were to mix zinc metal and COPPER ions in a container, this reaction would proceed by itself; we say that this reaction is SPONTANEOUS.

Answer:

Reduction Half REACTION E1/2 (V)

F2 + 2e− → 2 F− 2.87

Ce4+ + e− → Ce3+ 1.61

MnO4− + 8 H+ + 5e− → Mn2+ + 4 H2O 1.51

Cl2 + 2e− → 2Cl− 1.36

O2 + 4 H+ + 4e− → 2 H2O 1.23

Br2 + 2e− → 2 Br− 1.06

NO3− + 4 H+ + 3e− → NO + 2 H2O 0.96

AG+ + e− → Ag 0.80

Fe3+ + e− → Fe2+ 0.77

I2 + 2e− → 2I− 0.54

Cu2+ + 2e− → Cu 0.34

AgCl + e− → Ag + Cl− 0.222

Sn4+ + 2e− → Sn2+ 0.15

2H+ + 2e− → H2 0.000

Pb2+ + 2e− → PB −0.126

Ni2+ + 2e− → Ni −0.25

Cr3+ + e− → Cr2+ −0.41

Fe2+ + 2e− → Fe −0.44

Cr3+ + 3e− → Cr −0.74

Zn2+ + 2e− → Zn −0.76

Cr2+ + 2e− → Cr −0.91

Ba2+ + 2e− → Ba −1.57

Al3++ 3e− → Al −1.66

Mg2+ + 2e− → Mg −2.37

Na+ + e− → Na −2.714

Li+ + e− → Li −3.045

Answer:

Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

I2 + HNO3 → HIO3 + NO2 + H2O

Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking PLACE simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

I02 + H+1N+5O-23 → H+1I+5O-23 + N+4O-22 + H+12O-2

b) IDENTIFY and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number GOES down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.

O:

I02 → 2H+1I+5O-23 + 10e-

(I)

R:

H+1N+5O-23 + e- → N+4O-22

(N)

c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).

O:

I02 → 2H+1I+5O-23 + 10e-

R:

H+1N+5O-23 + e- → N+4O-22

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:

I02 → 2H+1I+5O-23 + 10e-

R:

H+1N+5O-23 + e- → N+4O-22

b) Balance the charge. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge.

O:

I02 → 2H+1I+5O-23 + 10e- + 10H+

R:

H+1N+5O-23 + e- + H+ → N+4O-22

c) Balance the oxygen atoms. CHECK if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:

I02 + 6H2O → 2H+1I+5O-23 + 10e- + 10H+

R:

H+1N+5O-23 + e- + H+ → N+4O-22 + H2O

Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E1/2 value.

Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:

I02 + 6H2O → 2H+1I+5O-23 + 10e- + 10H+

| *1

R:

H+1N+5O-23 + e- + H+ → N+4O-22 + H2O

| *10

O:

I02 + 6H2O → 2H+1I+5O-23 + 10e- + 10H+

R:

10H+1N+5O-23 + 10e- + 10H+ → 10N+4O-22 + 10H2O

Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

I02 + 10H+1N+5O-23 + 6H2O + 10e- + 10H+ → 2H+1I+5O-23 + 10N+4O-22 + 10e- + 10H2O + 10H+

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

I02 + 10H+1N+5O-23 → 2H+1I+5O-23 + 10N+4O-22 + 4H2O

Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.

Explanation:

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ANSWER:

I am SORRY I don't KNOW this answer because I did not UNDERSTAND this QUESTION

Answer:

determine the EMPIRICAL and molecular formula

of a COMPOUND composed of its 18.24g of C 0.51g of H and 16.91g of fluorine has a molar mass 562.0g/ml

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Answer:

C4H8 and C6H10

Explanation:TWO successive homologous MEMBERS differ by a difference of - CH2

ANSWER:

what is the heat ABSORBED or EVOLVED pls TELL

Answer:

Radioisotopes were USED for producing high yielding crop seeds to increase the AGRICULTURAL yield. ... Radiations from certain radioisotopes were also used for KILLING insects which damage the food grains. CEREALS, fruits, vegetables and canned food can be STORED for longer periods by gently exposing them to radiations.

Answer:

25.8658 g

Explanation:

Mass of SUBSTANCE = MOLES × Mr

Substitute the VALUES and u get the answer

EXPLANATION:

kis CHAPTER KA H ye BATAO