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Oxana [17]
3 years ago
14

Why is NaHCO3 a base?

Chemistry
1 answer:
worty [1.4K]3 years ago
7 0

NaHCO3 is a product of a strong base and a weak acid reaction. Thus it has weak basic properties.

HCO3- ion is actually amphoteric, which means it can act as a base or an acid. But it is weaker than a strong acid or a strong base.

<span>HCO3- is amphoteric meaning it acts both as a B.L. Acid and a B.L. Base.. which is why it's used to neutralize both acid and base spills in the lab.</span>

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A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of
Goryan [66]

Answer:

  • C₂ H₄ O

Explanation:

<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>

  • atomic mass of C: 12.0107 g/mol
  • molar mass of CO₂: 44.01 g/mol
  • Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
  • Solve for x:

         x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

<u />

<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>

  • atomic mass of H: 1.00784 g/mol
  • molar mass of H₂O: 18.01528 g/mol
  • proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
  • Solve for x:

        x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>

  • Mass of O = mass of pure sample - mass of C - mass of H
  • Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

<u>4) Mole calculations</u>

Divide the mass in grams of each element by its atomic mass:

  • C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
  • H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
  • O: 0.3840 g / 15.999 g/mol = 0.02400 mol

<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>

  • C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
  • H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
  • O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

  • <u>C₂ H₄ O </u>← answer
3 0
3 years ago
A student is doing an experiment that stoichiometry says should produce 34.6 grams of product. The student actually makes 25.2 g
attashe74 [19]
25.2 / 34.6 x 100
which would get you 73%

if you found my answer helpful pls give brainliest
6 0
2 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
Grace [21]

Answer:

0.21 g

Explanation:

The equation of the reaction is;

NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)

Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles

Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles

Since the reaction is 1:1, NaCl is the limiting reactant.

1 mole of NaCl yields 1 mole of AgCl

0.00147 moles of NaCl yields 0.00147 moles of AgCl

Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol

= 0.21 g

3 0
2 years ago
Earth’s cycles provide organisms with continued access to usable elements necessary for sustaining life. (True or False)
tensa zangetsu [6.8K]
The answer is truer
6 0
3 years ago
Read 2 more answers
Please help
aev [14]

Answer:

Copper(II) sulphate – sodium hydroxide reaction

The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.

5 0
3 years ago
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