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3 years ago

Please help I will mark brainliest if you answer first let's see who gets brainliest

Chemistry

2 answers:

shepuryov [24]3 years ago

8 0

Answer:

winter would be cooler and summers would be warmer

Explanation: i am hella late but......

Kitty [74]3 years ago

3 0

Answer:

Winter would be warmer, and summers would be cooler.

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A solution is prepared by dissolving 2.61 g ethanol (C2H5OH) in 16.2 g water. What is the molality (mol/kg) of C2H5OH in the sol

stiks02 [169]

Answer:

3.50 molal

Explanation:

Molality → Moles of solute / kg of solvent.

Let's convert the solvent's mass from g to kg

16.2 g . 1kg / 1000 g = 0.0162 kg

Let's determine the moles from the solute

2.61 g . 1 mol / 46 g = 0.0567 moles

Molality → 0.0567 mol / 0.0162 kg = 3.50 m

7 0

3 years ago

Explain why an indicator is used in titration

stellarik [79]

Answer:

Titrations. Because a noticeable pH change occurs near the equivalence point of acid-base titrations, an indicator can be used to signal the end of a titration. When selecting an indicator for acid-base titrations, choose an indicator whose pH range falls within the pH change of the reaction.

Hope it helped!!

4 0

2 years ago

The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p

earnstyle [38]

Answer:

The frequency is $f = 2,01 * 10^{15} \ Hz$

Explanation:

From the question we are told that

The energy required to ionize boron is $E_b = 801 KJ/mol$

Generally the ionization energy of boron pre atom is mathematically represented as

$E_a = \frac{E_b}{N_A}$

Here $N_A$ is the Avogadro's constant with value $N_A = 6.022*10^{23}$

$E_a = \frac{801}{6.022*10^{23}}$

=> $E_a = 1.330 *10^{-18} \ J/atom$

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

$E = hf = E_a$

=> $hf = E_a$

Here h is the Planks constant with value $h = 6.626 *10^{-34}$

$f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}$

=> $f = 2,01 * 10^{15} \ Hz$

8 0

2 years ago

Comparison between alkyl nucleophilic substitution with acyl nucleophilic substitution

S_A_V [24]

Answer:

in both nucleophil attach the c and leaving group leave but in acyl nu. subsituation c of carbonyl because of double bond with o have bigger positive charge and is better electrophil so do it faster,also alkyl nu. subsituation can have rearangment if going from sn1 and in sn2 sterichemistry of molecule change , acyl nu. subsituation most of time is better

4 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago