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Lubov Fominskaja [6]
2 years ago
8

Please help me with this question....! Question in attachment

Chemistry
2 answers:
andrezito [222]2 years ago
6 0
2-chloro-5-ethylhexane

This is an alkane hydrocarbon with an attached halogen and ethyl branch

The chlorine takes priority over the ethyl branch when naming
Phoenix [80]2 years ago
6 0
I think it’s branch when naming I am not sure..
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Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How ma
Art [367]

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

3 0
2 years ago
What is the molarity of 0.26 mol of H2SO4 dissolved in 0.3 L of solution? *
irina1246 [14]

Answer:

A

Explanation:

molarity=moles of solute/liter of solution

molarity=0.26/0.3

molarity=0.87molar

7 0
3 years ago
Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;
marishachu [46]

<u>Answer:</u> Pairs are:  (a) and (d), (b) and (f), (c) and (e)

<u>Explanation:</u>

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

  • <u>For a:</u>

The given electronic configuration is:  1s^22s^22p^63s^2

The number of valence electrons in the given configuration are 2

  • <u>For b:</u>

The given electronic configuration is:  1s^22s^22p^63s^3

The number of valence electrons in the given configuration are [2 + 3] = 5

  • <u>For c:</u>

The given electronic configuration is:  1s^22s^22p^63s^23p^64s^23d^{10}4p^6

The number of valence electrons in the given configuration are [2 + 6] = 8

  • <u>For d:</u>

The given electronic configuration is:  1s^22s^2

The number of valence electrons in the given configuration are 2

  • <u>For e:</u>

The given electronic configuration is:  1s^22s^22p^6

The number of valence electrons in the given configuration are [2 + 6] = 8

  • <u>For f:</u>

The given electronic configuration is:  1s^22s^22p^63s^23p^3

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are:  (a) and (d), (b) and (f), (c) and (e)

4 0
3 years ago
A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?
SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar  mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M =  0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K /  1 atm

M = 20.5 g/mol

8 0
2 years ago
A solution contains 45 grams of nitrogen gas and 40 grams of argon. What is the mole fraction of argon in this solution?
Katarina [22]
 <span>0.38 

You first calculate the total moles by dividing the grams by molecular weight: 
45 g N2 / 28.02 g/mol = 1.6 mol N2 
40 g Ar / 39.95 g/mol = 1.0 mol 

Then you divide the moles of Ar by the total number of moles: 
1.0 / (1.6 + 1.0) = 0.38 mol fraction</span>
8 0
3 years ago
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