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Dennis_Churaev [7]
3 years ago
6

On the celsius scale, at what temperature does water boil?

Chemistry
1 answer:
Gnom [1K]3 years ago
6 0
Simple....

On the celsius scale, at what temperature does water boil? 

---->>>

D.)100

Thus, your answer.

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How is the food in a stomach broken down into simpler substances? What chemicals help this process?
Ksenya-84 [330]

Answer:

with the help of the juice contained in it

4 0
3 years ago
Which compound does not form a precipitate when reacted with sodium hydroxide?
Alla [95]

Answer:

NaNO₃

Explanation:

A precipitate is a compound or a salt formed from a precipitation reaction and does not dissolve in water and therefore will exist in solid state.

From the choices given precipitation reaction will occur between;

  • Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq)
  • Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)
  • FeSO₄(aq) + 2NaOH(aq) → Fe(OH)₂(s) + Na₂SO₄(aq)

Fe(OH)₃, Cu(OH)₂, and Fe(OH)₂ are precipitates.

From the rules of solubility, hydroxides are insoluble except Ca(OH)₂ which is slightly soluble and hydroxides of ammonium and alkali metals.

5 0
3 years ago
What volume of oxygen is needed to react with 20cm3 of ethane?
Anna71 [15]

Answer:

The reaction states that 2 moles of ethane react with 7 moles of oxygen. At standard temperature and pressure conditions (STP), each mole of gas occupies a volume of 22.4 liters. Therefore, 2 moles of ethane occupy liters, and 7 moles of oxygen occupy liters. In other words: 6 liters.

Explanation:

I hope it helps you

7 0
3 years ago
). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
Nikitich [7]

Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

7 0
3 years ago
The melting point of the alkali metals (increased/decreases) from lithium to francium
Lynna [10]
The melting point would decrease
3 0
3 years ago
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