Answer: Air, sea water, and carbonation dissolved in soda are all examples of homogeneous mixtures, or solutions. Hope this helps :)
M ( HCl ) = ?
V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L
M ( NaOH ) = 0.113 M
V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L
number of moles NaOH:
n = M x V
n = 0.113 x <span> 0.0512 => 0.0057856 moles of NaOH
mole ratio:
</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V
M = 0.0057856 / <span>0.0255
</span>
= 0.227 M
Answer A
hope this helps!
<span>1.61 × 1023 Multiply by 26.8 to get the answer.161.33 x 10 ^23 </span>
<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 × 
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Given data:
Moles of hydrochloric acid = 0.385 mol
Mass of chlorine gas =?
Chemical equation:
4HCl + O₂ → 2Cl₂ + 2H₂O
Now we will compare the moles of Cl₂ with HCl.
HCl : Cl₂
4 : 2
0.385 : 2÷4× 0.385 = 0.1925 mol
Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.
Mass of Cl₂ :
Mass of Cl₂ = moles × molar mass
Mass of Cl₂ =0.1925 mol × 71 g/mol
Mass of Cl₂ = 13.6675 g
Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
Learn more about moles here:
brainly.com/question/8455949
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