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2 years ago

7. In a remote area of Columbia, kids have to travel to school on a zip-line!

Chemistry

1 answer:

andriy [413]2 years ago

6 0

Answer:

yes

Explanation:

Kids like fun things, and school is not fun , so if kids got to go to school on a zipline, attendance would be through the roof!

You might be interested in

Which postulate of Dalton's atomic model was later changed and why?

telo118 [61]

Answer:

Explanation:

Part two of Dalton's theory had to be modified after mass spectrometry experiments demonstrated that atoms of the same element can have different masses because the number of neutrons can vary for different isotopes of the same element. ... Scientists have even developed the technology to see the world on an atomic level!

hoped i helped you :)

4 0

2 years ago

A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula

Sever21 [200]

Answer:

Na₂CO₃•H₂O

Explanation:

After it is heated, the remaining mass is the mass of sodium carbonate.

30.2 g Na₂CO₃

Mass is conserved, so the difference is the mass of the water:

35.4 g − 30.2 g = 5.2 g H₂O

Convert masses to moles:

30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃

5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O

Normalize by dividing by the smallest:

0.285 / 0.285 = 1.00 mol Na₂CO₃

0.289 / 0.285 = 1.01 mol H₂O

The ratio is approximately 1:1. So the formula of the hydrate is Na₂CO₃•H₂O.

3 0

2 years ago

Read 2 more answers

How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) acc

Novay_Z [31]

Answer:

429.4 kJ are absorbed in the endothermic reaction.

Explanation:

The balanced chemical equation tells us that 1168 kJ of heat are absorbed in the reaction when 4 mol of NH₃ (g) react with 5 mol O₂ (g).

So what we need is to calculates how many moles represent 25 g NH₃(g) and calculate the heat absorbed. (NH₃ is the limiting reagent)

Molar Mass NH₃ = 17.03 g/mol

mol NH₃ = 25.00 g/ 17.03 g/mol = 1.47 mol

1168 kJ /4 mol NH₃ x 1.47 mol NH₃ = 429.4 kJ

6 0

2 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

What is the mass percent of Fe in iron(II) chloride

GREYUIT [131]

Its

28.090%
hope its helpful >~

4 0

1 year ago