Answer:
0.89
Explanation:
You are mixing an acid and a base, so there will be a neutralization reaction.
One of the reactants will be in excess, so we must determine its concentration and then calculate the pH.
<em>Moles of NaOH
</em>
Moles of NaOH = 25 × 0.173
Moles of NaOH = 4.32 mmol
===============
<em>Moles of HCl
</em>
Moles of HCl = 35 × 0.342
Moles of HCl = 12.0 mmol
===============
<em>Amount of excess reactant
</em>
NaOH + HCl ⟶ NaCl +H₂O
<em>n</em>/mmol: 4.32 12.0
The 4.32 mmol of NaOH reacts completely with 4.32 mmol of HCl.
Excess HCl = 12.0 – 4.32
Excess HCl = 7.6 mmol
===============
<em>Concentration of the excess HCl
</em>
Total volume = 25 + 35
Total volume = 60 mL
<em>c </em>= millimoles HCl/millilitres HCl
<em>c</em> = 7.6/60
<em>c</em> = 0.13 mol/L
===============
<em>Calculate the pH
</em>
The HCl dissociates completely to hydronium ions, so
[H₃O⁺] = 0.13 mol·L⁻¹
pH = -log[H₃O⁺]
pH = -log0.13
pH = 0.89
Explanation:
There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.
Answer:
The answer to your question is: 85.458 amu
Explanation:
data
Rubidium-85 A = 84.9118 amu abundance = 72.15%
Rubidium - 87 A = 86.9092 amu abundance = 27.85%
Atomic weight = ?
Atomic weight = 84.9118(0.7215) + 86.9092(0.2785)
Atomic weight = 61.2538 + 24.2042
Atomic weight = 85.458 amu
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
brainly.com/question/23580857
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