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3 years ago

A bond between a nonmetal and a nonmetal is called a(n)

Chemistry

2 answers:

Blababa [14]3 years ago

8 0

B is the answer I think

Pani-rosa [81]3 years ago

7 0

Answer:

answer: B

Explanation:

becaucse it is B

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Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a

kiruha [24]

From the calculation, the molar mass of the solution is 141 g/mol.

<h3>What is the molar mass?</h3>

We know that;

ΔT = K m i

K = the freezing constant

m = molality of the solution

i = the Van't Hoft factor

The molality of the solution is obtained from;

m = ΔT/K i

m = 3.89/5.12 * 1

m = 0.76 m

Now;

0.76 = 26.7 /MM/0.250

0.76 = 26.7 /0.250MM

0.76 * 0.250MM = 26.7

MM= 26.7/0.76 * 0.250

MM = 141 g/mol

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5 0

2 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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2 years ago

As the ph approaches 0, what happens to the concentration of h3o+ ions?

natta225 [31]

We know that

pH = -log[H+]

the pH value falls in between 0- 7 for acids

As the pH value increases the concentration of [H+] increases.

similarly as the value of pH approaches 0, the concentration of H+ increases

The solution said to become more acidic

Also

[H+] X [OH-] = 10^-14

Thus pH + pOH = 14

hence the concentration of OH- decreases as the pH approaches zero

6 0

3 years ago

A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 m

DerKrebs [107]

The pH of the solution after the addition of 100. 0 ml of HNO3. The kb of NH3 is 1. 8 × 10-5. So, pH is 10.9 basic .

This neutralization occurs as the acid is added to the base:

NH3(aq)+ HNO3(aq)= NH4NO3(aq)+ H2O(l)

The initial moles of NH3present is given by,

nNH3= c×v= 0.10× 100/1000= 0.01m

The number of moles of

HNO3 added is given by:

nHNO3= c×v= 0.10× 100/1000= 0.001

It is clear from the equation that the acid and base react in a 1:1 molar ratio. So, the no. moles of NH3remaining will be 0.01 - 0.001= 0.009

The total volume is now

100.0+100.0= 200.0x cm3

The concentration of NH3 is given by,

[NH3]= c/v= 0.001/200/1000= 0.05x mol/l .

pOH= 12 (pKb− logb)

where b is the base's concentration.

Given that there is little dissociation, we can roughly compare this to the starting concentration.

pKb= −logKb= −log(1.8×10−5)= 4.744

pOH= 3.056

At 25∘xC, we know that, pH+ pOH=14

pH= 14− 3.056= 10.9

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8 0

2 years ago

Which statement correctly describes the relationship between reactant and yield?

wlad13 [49]

Answer:

The theoretical yield is calculated from the amount of the limiting reactant present.

3 0

3 years ago