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Margaret [11]
3 years ago
5

What mass of hydrochloric acid (in grams) can 2.7 g of sodium bicarbonate neutralize? (Hint: Begin by writing a balanced equatio

n for the reaction between aqueous sodium bicarbonate and aqueous hydrochloric acid.)
Chemistry
1 answer:
Julli [10]3 years ago
3 0

Answer:

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

Explanation:

Step 1: Data given

Mass of sodium bicarbonate = 2.7 grams

Step 2: The balanced equation

HCl + NaHCO3 ⇔  NaCl + H2O + CO2

Step 3: Calculate moles NaHCO3

moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles

Step 4: Calculate moles HCl

For 1 mol NaHCO3 we need 1 mol HCl

For 0.032 moles NaHCO3 = 0.032 moles HCl

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass HCl

mass HCl = 0.032 * 36.46 g/mol= 1.17 grams

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

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4 Al + 3 O2 --> 2 Al2O3
SIZIF [17.4K]

Answer:

The answer to your question is 7.4 moles of Aluminum

Explanation:

Data

moles of Al = ?

moles of Al₂O₃ = 3.7

Balanced chemical reaction

                4 Al  +  3 O₂  ⇒  2 Al₂O₃

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.

                4 moles of Aluminum ----------------- 2 moles of Al₂O₃

                 x                                  ----------------- 3.7 moles of Al₂O₃

                            x = (3.7 x 4) / 2

                            x = 14.8 / 2

                            x = 7.4 moles of Aluminum

5 0
3 years ago
State the postulate of Bohr theory
KengaRu [80]

Answer:

Bohr's model of the hydrogen atom is based on three postulates:

1) An electron moves around the nucleus in a circular orbit,

2) An electron's angular momentum in the orbit is quantised,

3) The change in an electron's energy as it makes a quantum jump from one orbit to another is always accompanied by the emission or absorption of a photon. Bohr's model is semi-classical because it combines the classical concept of electron orbit (postulate 1) with the new concept of quantisation ( postulates 2 and ).

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3 years ago
If a bar magnet's neutral region is broken into two, what will most likely occur? A. Neither segment will have a north or south
grigory [225]
The answer is D I JUST TOOKM THE TEST
8 0
3 years ago
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What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?
Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>

<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>

<em>∴ The freezing point of the solution is - 4.39 °C.</em>

6 0
3 years ago
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bixtya [17]
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3 0
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