8. The threats to biodiversity are except:.

For the hypothetical compound AX2, which of the following statements is true?. a.) A is a nonmetal from Group 3A, and X is a met

Ierofanga [76]

The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X belonging to Group 7A with -1 charge. Answer is D.

4 0

3 years ago

Read 2 more answers

Which of these statements is incorrect regarding an atom of the element scandium?

UkoKoshka [18]

C is incorrect. I has 21 PROTONS in its nucleus. Not neutrons.

4 0

3 years ago

Treatment of ethyl acetoacetate with NaOEt (2 equiv) and BrCH2CH2Br forms compound X. This reaction is the first step in the syn

valkas [14]

Answer:

See explanation below

Explanation:

In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.

In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.

The mechanism of this reaction to reach X is shown in the attached picture.

3 0

4 years ago

Questions 3,4,5<br> Brainlist

Diano4ka-milaya [45]

Answer:

3. 126.02 g of H₂S and 1109.2 g of NaI

4. 364.5g of Mg

5. 361.08 g of MgSO₄ and 108 g of H₂O

Explanation:

3. This is the reaction

2HI + Na₂S → H₂S + 2NaI

7.4 moles of HI react, so, ratio is 2:1 with H₂S and 2:2 with NaI

We would produce the same moles of NaI, 7.4 moles and the half of moles, of H₂S (7.4 /2) = 3.7 moles

Let's convert the moles to mass (mol . molar mass)

3.7 mol H₂S . 34.06 g/mol = 126.02 g

7.4 mol NaI . 149.9 g/mol = 1109.2 g

4. The balanced reaction is this:

3Mg + N₂ → Mg₃N₂

Ratio is 1:3. Therefore 1 mol of Mg₃N₂ were produced by 3 moles of Mg.

So, 5 moles of Mg₃N₂ would have been produced by 15 moles of Mg. (5 .3)

Let's convert the moles to mass (moles . molar mass)

15 mol . 24.30 g/mol = 364.5 g

5. The reaction is: H₂SO₄ + Mg(OH)₂ → MgSO₄ + 2H₂O

Ratios are 1:1 and 1:2, between sulfuric acid and the products.

If I have 3 moles of acid, I would produce 3 moles of magnessium sulfate and 6 moles of water (3 .2)

Let's convert the moles to mass

3 mol . 120.36 g/mol = 361.08 g of MgSO₄

6 mol . 18 g/mol = 108 g of H₂O

8 0

4 years ago

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0

3 years ago