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2 years ago

A____molecule is an example of a carbohydrate

Chemistry

1 answer:

denpristay [2]2 years ago

5 0

I think it’s CTell me if I’m wrong

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A trisubstituted cyclohexane compound is given below in its chair conformation. Draw the corresponding planar (overhead) represe

REY [17]

Answer:

See explaination

Explanation:

Chiral carbons are carbons that have four different groups, so C atoms 1, 4, and 5 are chiral carbons and thus H atoms are shown for those atoms.

6 0

2 years ago

Group 2 in the penodic table contains the elements beryllium (Be), magnesium (Mg) , calcium (Ca) , strontium (Sf) )barium (Ba)an

mario62 [17]

Answer:

They are all alkali earth metals.

Explanation:

Their valence shell each has 2 electrons. Also, they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure. They form alkaline solutions, hydroxides, when reacting with water and their oxides are found in the earth’s crust.

6 0

2 years ago

Which element in group 2 (Alkaline Earth Metals) is the most reactive?

never [62]

Answer:

the reactivity of alkaline earth metals increases from top to the bottom of the group, that's because the atoms get bigger from the top to the bottom so the valence electrons are farther from the nucleus.

8 0

2 years ago

Calculate the PH of a solution 0.030 MH2SO4

Zinaida [17]

Answer:

pH= 2- log3

Explanation:

H2SO4 + H2O -> HSO4^(-) + H30^(+)

0.03M ___ ___

___ 0.03M 0.03M

H30^(+) : C = 0.03M

pH= - log( [H3O^(+)] ) => pH= - log {3× 10^(-2)} => pH = 2 - log3

4 0

3 years ago

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = 0,0989 moles NaOH

Moles of Al₂(SO₄)₃ are:

15,8g × $\frac{1mol}{342,15g}$ = 0,0462 moles Al₂(SO₄)₃

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × $\frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH}$ = 0,0165 moles Al₂(SO₄)₃

As you have 0,0462 moles Al₂(SO₄)₃ the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × $\frac{2molAl(OH)_{3}}{6 moles NaOH}$ = 0,0330 moles of Al(OH)₃

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = 2,57 g of Al(OH)₃ ≡ mass of precipitate

I hope it helps!

3 0

3 years ago