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2 years ago

After each step in a dichotomous key, the questions get ________.

Chemistry

2 answers:

erica [24]2 years ago

8 0

I think it’s the 1 one lmk if it’s correct

GenaCL600 [577]2 years ago

4 0

Answer:

I believe the answer is 1. more specific.

Explanation:

Since the dichotomous key is used to identify a species, you would have to get down to the specifics, so specificity is required.

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Explain why 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane. hint: examine the

Mama L [17]

Answer:-

The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.

The I - attacks from backside to give the transition state for both.

If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.

This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.

7 0

3 years ago

Stoichiometry: Calculate the moles of H2 produced by .042 g Mg in the equation Mg + 2HCl > MgCl2 + H2

Evgesh-ka [11]

Answer:

<em>Mg </em>(<em>s</em>) + 2<em>HCI2 </em>(<em>aq</em>) → <em>MgCI2 </em>(<em>aq</em>) + <em>H2 </em>(<em>g</em>)

I think this is the correct answer I not a 100% sure if it is correct.

Explanation:

Guessing

6 0

3 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

3 years ago

The equilibrium constant K changes with changes in<br>the temperature.

chubhunter [2.5K]

<h3>Answer:</h3><h2>Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. You can see that as the temperature increases, the value of Kp falls.</h2>

4 0

2 years ago

Write the symbol for each of the following ions:

kap26 [50]

Answer:

a) $_{31}^{71}\textrm {P^{3+}}$

b) $_{35}^{80}\textrm {Br^{-}}$

c) $_{90}^{232}\textrm {Th^{4+}}$

d) $_{38}^{87}\textrm {Sr^{2+}}$

Explanation:

For neutral atoms:

Atomic Number (Z)= number of protons = number of electrons

Mass number (A) = number of protons + number of neutrons

For ions with positive net charge:

Number of protons = Z + net charge

For ions with negative net charge:

Number of protons= Z - net charge

a) A = 71, Charge = +3

Number of electrons = 28

Number of protons = 28 +3 =31

$_{31}^{71}\textrm {P^{3+}}$

b) A = 35, Z = 45+35=80, Charge = -1

Number of protons =35

Number of neutrons = 45

Number of electrons = 36

Charge = Number of protons- Number of electrons =35-36 = -1

$_{35}^{80}\textrm {Br^{-}}$

c) Charge = +4

Number of electrons = 86

Number of protons = Z = 86+4 = 90

mass number = A = 90+142 = 232

$_{90}^{232}\textrm {Th^{4+}}$

d) Charge = +2

Atomic number = Number of protons = Z = 38

mass number = A = 87

$_{38}^{87}\textrm {Sr^{2+}}$

3 0

3 years ago