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Marina86 [1]
3 years ago
9

Jeremy had a set of 500 toy blocks that could be connected by snapping them together. At first, he used all the blocks to build

a small house. Later, he rearranged all the blocks to build a small car. How did the mass of the house compare to the car?
Chemistry
2 answers:
valentina_108 [34]3 years ago
6 0
The masses were the same
Nat2105 [25]3 years ago
3 0

Answer:

The masses were the same.

Explanation:

study isalnd :)

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How many grams of barium sulfate, baso4, are produced if 25.34 ml of 0.113 m bacl2 completely react given the reaction: bacl2 +
dimulka [17.4K]
Answer is: mass of barium sulfate is 0.668 grams.
Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄<span> + 2NaCl.
V(</span>BaCl₂) = 25.34 mL ÷ 1000 mL/L = 0.02534 L.
c(BaCl₂) = 0.113 mol/L.
n(BaCl₂) = V(BaCl₂) · c(BaCl₂).
n(BaCl₂) = 0.02534 L · 0.113 mol/L.
n(BaCl₂) = 0.00286 mol.
From chemical reaction: n(BaCl₂) : n(BaSO₄) = 1 : 1.
n(BaSO₄) = 0.00286 mol.
m(BaSO₄) = n(BaSO₄) · M(BaSO₄).
m(BaSO₄) = 0.00286 mol · 233.4 g/mol.
m(BaSO₄) = 0.668 g.
8 0
3 years ago
Read 2 more answers
Which of the following are saturated hydrocarbons?
Digiron [165]

Answer:

a. Alkynes (CnH2n+2) are saturated hydrocarbons.

6 0
2 years ago
. How many liters of NH3, at STP, will react with 7.4g O₂?
iren [92.7K]

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Explanation:

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3 0
2 years ago
Please help!! ASAP Calculate the pH of the solution after the addition of the following amounts of 0.0574 M HNO3 to a 80.0 ml so
alexira [117]

Answer:

a) 10.457.

b) 9.32.

c) 8.04.

d) 6.58.

e) 4.76.

f) 2.87.

Explanation:

  • Aziridine is an organic compounds containing the aziridine functional group, a three-membered heterocycle with one amine group (-NH-) and two methylene bridges (-CH2-). The parent compound is aziridine (or ethylene imine), with molecular formula C2H5N.
  • Aziridine has a basic character.
  • It has pKa = 8.04.
  • So, pKb = 14 – 8.04 = 5.96
  • Kb = 1.1 x 10⁻⁶.
  • If we denote Aziridine the symbol (Az),  It is dissociated in water as:

Az + H₂O → AzH⁺ + OH⁻

<u><em>a) 0.00 ml of HNO₃: </em></u>

There is only Az,

[OH⁻] = √(Kb.C)

Kb = 1.1 x 10⁻⁶. & C = 0.0750 M.

[OH⁻] = √(1.1 x 10⁻⁶)(0.075) = 2.867 x 10⁻⁴.

∵ pOH = - log[OH-] = - log (2.867 x 10⁻⁴) = 3.542.

∴ pH = 14 – pOH = 14 – 3.542 = 10.457.

<u><em>b) 5.27 ml of HNO₃</em></u>

  • To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
  • No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
  • No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (5.27 ml) = 0.302 mmol.
  • The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (0.302).
  • This will form a basic buffer in the presence of weak base (Az).

<em>pOH = pKb + log[salt]/[base] </em>

  • [salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (0.302) / (85.27) = 3.54 x 10⁻³ M.
  • [base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 0.302 mmol) / (85.27 ml) = 0.0668 M.
  • pOH = pKb + log[salt]/[base] = 5.96 + log[3.54 x 10⁻³]/[ 0.0668 M] = 4.68.
  • <em>pH = 14 – pOH = 14 – 4.68 = 9.32. </em>

<em />

<em><u>c) Volume of HNO₃ equal to half the equivalence point volume :</u></em>

  • At half equivalence point, the concentration of the salt formed is equal to the concentration of the remaining base (aziridine), [salt] = [base].
  • pOH = pKb + log[salt]/[base] = 5.96 + log[1.0] = 5.96.
  • pH = 14 – pOH = 14 – 5.96 = 8.04.

<u><em>d) 101 ml of HNO₃: </em></u>

  • To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
  • No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
  • No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (101.0 ml) = 5.7974 mmol.
  • The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (5.7974).
  • This will form a basic buffer in the presence of weak base (Az).

<em>pOH = pKb + log[salt]/[base] </em>

  • [salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (5.7974) / (181.0) = 0.032 M.
  • [base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 5.7974 mmol) / (181.0 ml) = 0.00112 M.
  • pOH = pKb + log[salt]/[base] = 5.96 + log[0.032]/[ 0.00112] = 7.416.
  • pH = 14 – pOH = 14 – 7.416 = 6.58.

<u><em>e) Volume of HNO₃ equal to the equivalence point :</em></u>

  • At the equivalence point the no. of millimoles of the base is equal to that of the acid.
  • Volume of HNO₃ needed for the equivalence point = (6.00 mmol) / (0.0574 mmol/ml) = 104.5 ml  

At the equivalence point:  

  • [Az] = 0.
  • [AzH⁺] = (6.00 mmol) / (80.0 + 104.5 ml) = 0.0325 M.
  • As Ka is very small, the dissociation of AzH⁺ can be negligible.  

Hence, [AzH⁺] at eqm ≈ 0.0325 M.

  • [H+] = √(Ka.C) = √(10⁻⁸˙⁰⁴ x 0.0325) = 1.72 x 10⁻⁵.
  • pH = - log[H+] = - log(1.72 x 10⁻⁵) = 4.76.

<u><em>f) 109 ml of HNO₃: </em></u>

  • No. of milli-moles of H⁺ added from HNO₃ = (0.0574 mmol/ml) × (109 ml) = 6.257 mmol.
  • Which is higher than the no. of millimoles of the base (Az) = 6.0 mmol.
  • After the addition, [H⁺] = (6.257 - 6.00) / (80.0 + 109 mL) = 0.00136 M.
  • As Ka is small and due to the common ion effect in the presence of H⁺, the dissociation of Az is negligible.  
  • pH = -log[H⁺] = -log(0.00136) = 2.87.
3 0
3 years ago
ANSWER ASAP PLEASE!<br> Balance the following equation:<br> _Fe + _O₂ --&gt; _FeO₃
ira [324]
It would be. 2Fe + 3O2 ——> 2Fe03. You have to make oxygen 6 on both sides in order to balance, then you can do Fe. Which on the right would be 2 by making oxygen 6. Then add the 2 to the other side to balance it.
8 0
2 years ago
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